Does time dilation work in 1d space?

[YouAreAwesome]

TL;DR Summary

Imagine that Alice is stationary and Bob is travelling at the speed of light away from her to the right. If both Alice and Bob send a light signal towards the other that rotates through the colours of the rainbow each second, and after 10 seconds Bob returns, what are the consequences?

Imagine this question in 2 dimensions, time (t) and distance (x), that is (t,x). Alice (A) is at the origin, x=0. Bob (B) begins at x=c. Thus we have A(0,0) and B(0,c). Both Alice and Bob send a light signal towards the other but let's say the signal changes colour every second by the colours of the rainbow (i.e. the first is red, the second is orange using ROYGBIV and continues to rotate through the colours). When t=0 the experiment begins and each send a red light towards each other. But when t=1 Bob starts moving to the right at the speed of light (c). Alice remains stationary throughout the experiment. At t=10 Bob stops moving right and returns to the left finishing his journey at the origin.

From Bob's frame of reference he only sees Alice's first red pulse of light for the first 9 seconds (from t=1 to t=10). It would seem as though time has stopped for Alice because she's only sent the one signal at t=0 that he received at t=1 and continues to see for the next 9 seconds. He then turns back towards the origin and what does he see on his way back? After 1 second he would see the Orange pulse. That is B(11,9c)=Orange. So I'll write this out in a table.

BOB'S POSITION (t,x)	BOB SEES	ALICE'S POSITION (t,0)	ALICE SEES
B(0,c)	Nothing	No need to write all these out	Nothing
B(1,c)	Red		Red
B(2,2c)	Red		Orange
B(3,3c)	Red		?
B(4,4c)	Red
B(5,5c)	Red
B(6,6c)	Red
B(7,7c)	Red
B(8,8c)	Red
B(9,9c)	Red
B(10,10c)	Red
B(11,9c)	Orange
B(12,8c)	Yellow
B(13,7c)	Green
B(14,6c)	Blue
B(15,5c)	Indigo
B(16,4c)	Violet
B(17,3c)	Red
B(18,2c)	Orange
B(19,c)	Yellow
B(20,0)	Indigo

It seems while Bob experiences a normal 20 seconds of time, he only sees 10 changes of light from Alice, which corresponds to 10 seconds passing for Alice.

First, is this just plain wrong for some reason?
Second, what does Alice see?

I am new to relativity so please excuse any ignorance or misunderstanding, I'm here to learn!

Thanks for any replies.

 

[Nugatory]

You cannot construct a consistent thought experiment if you assume that Bob is moving at speed $c$ because it’s not possible to travel at the speed of light. However, we can fix that by changing the problem so that Bob is moving at some speed less than $c$; you will find that either $.6c$ or $.8c$ make the calculation easier. And while we’re simplifying things, let’s measure distances in light-seconds so that $c=1$ (light speed is one light-second per second) so Bob’s initial position is (0,1).

Another complication comes from using colored light. When the two are moving apart the light one emits will be redshifted when received by the other; when moving towards one another there will be a blueshift. So Alice won’t see red light coming from Bob at the end of the first second, she won’t see anything because the red light Bob emitted will be redshifted down into the infra-red. Alice can compensate for this by calculating effects of Doppler, but it’s easier to have Bob send a video stream of his clock - the image of the clock changes instead of the color of the light changing.

But with these tweaks, you can work out what happens. You have to allow for the effects of relativity of simultaneity and time dilation but you will end up with a internally consistent scenario in which less time passes for Bob during his roundtrip then for Alice. In fact, this is a variation of the so-called Twin Paradox, extensively discussed here and elsewhere.

You will want to read http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html and pay particular attention to the section on the Doppler explanation. It would also be a really good exercise to try drawing a Minkowski diagram of the whole thing, showing the paths of the light signals between Alice and Bob.

[edit: fixed broken link]

 

[PeroK]

I am new to relativity so please excuse any ignorance or misunderstanding, I'm here to learn!

You could try the first chapter of Morin's book here:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

 

[Mister T]

You can also take a look at the YouTube video (search for Paul Hewitt, Twin Trip). The scenario shown there is very close to what you describe it's just that light pules are used instead of light colors, but the same logic applies to both.

 

[YouAreAwesome]

You could try the first chapter of Morin's book here:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

Thanks for the link to the book chapter, I'm making my way through it slowly and I'm a little confused on this point the author makes regarding the flash of light from the centre of a moving train to receivers on either end, with B observing from outside:

Returning to our setup with the light beams and receivers, we can say that because B sees both light beams move with speed c, the relative speed (as viewed by B) of the light and the left receiver is c + v, and the relative speed (as viewed by B) of the light and the right receiver is c − v.
Remark: Yes, it is legal to simply add or subtract these speeds to obtain the relative speeds as viewed by B.

I thought this was illegal. I thought if we measure the speed of light in the positive x direction relative to the speed of light in the negative x direction we are not allowed to attain a velocity of 2c. i.e. c - (-c) = 2c. I was of the understanding the fastest velocity is c, as in, it is a maximum. And that 2c is impossible even in any relative scenario. In the example given, if we are allowed to say light can travel relative to B at the speed of c+v, then doesn't that break the principle:

The speed of light in vacuum has the same value c (approximately3 · 108 m/s) in any inertial frame.

If the inertial frame is at B (does it make sense to say this??) then light is traveling at c+v relative to B. How is this not a problem?

 

[Ibix]

Remark: Yes, it is legal to simply add or subtract these speeds to obtain the relative speeds as viewed by B.

I thought this was illegal.

If I've got an object A going at 0.5c in one direction and an object B going at 0.5c in the opposite direction then the distance between them as measured by me grows at c. That's what Morin is talking about. That does not mean that A is moving away from B at c as measured by B. That's what you are talking about and you are correct that this is always less than c.

If the inertial frame is at B (does it make sense to say this??) then light is traveling at c+v relative to B. How is this not a problem?

You mean, I think in the inertial frame where B is at rest then light is traveling at c+v. This isn't correct - see above.

The point is that there are two separate concepts that I usually call "relative speed" and "separation rate". I, working in an inertial frame where I'm at rest, never see anything traveling faster than c. That is, nothing has a relative speed greater than c. But that means that the distance between two things traveling in opposite directions can grow at up to 2c - that's the separation rate.

Now going back to my example of two objects traveling in opposite directions at 0.5c - I see them separating at c. But (spoiler!) I measure their rulers length contracted and their clocks time dilated. I also measure their clocks to be out of sync if they aren't right next to each other - so I am unsurprised that they don't measure the other object's speed to be c. Their clocks and rulers aren't measuring the same things mine are.

Each object sees its own clocks and rulers behaving normally and sees mine and the other object's behaving oddly. It all comes out in the maths to be consistent, but it's worth keeping in mind this distinction between speed and separation rate - they're the same thing in Newtonian physics, but not in relativity.

The other thing to bring to your attention is the point I made in passing about measuring moving clocks to be out of sync. This is called the relativity of simultaneity, and probably 99% of failures to understand relativity come from not realising it exists. It's not as famous as time dilation and length contraction, but probably more important. I won't go into it here, but I wanted to warn you to remember to figure it into your thinking - relativity becomes paradoxical very quickly if you aren't aware of it.

 

[PeroK]

Thanks for the link to the book chapter, I'm making my way through it slowly and I'm a little confused on this point the author makes regarding the flash of light from the centre of a moving train to receivers on either end, with B observing from outside:

Returning to our setup with the light beams and receivers, we can say that because B sees both light beams move with speed c, the relative speed (as viewed by B) of the light and the left receiver is c + v, and the relative speed (as viewed by B) of the light and the right receiver is c − v.
Remark: Yes, it is legal to simply add or subtract these speeds to obtain the relative speeds as viewed by B.

I thought this was illegal.

If the inertial frame is at B (does it make sense to say this??) then light is traveling at c+v relative to B. How is this not a problem?

It's useful to distinguish here between relative speed and separation speed. In this case Morin is talking about separation speed. Tha maximum separation speed is $2c$. You can see this as follows:

Assume an object on you left is moving towards you at a speed close to $c$, and another object on your right is also moving towards you at close to $c$. By the definition of speed, both objects cover nearly $300,000 km$ in $1s$. Therefore, the separation between them is, by definition, reducing at nearly $600,000 km/s$. In your reference frame.

In fact, if all velocities are measured in one reference frame, then the usual rules of vector addition apply to give separation velocities.

If, however, you switch to the reference frame of one of those objects, then the other object cannot be moving towards it at nearly $2c$. Instead, the relative speed in the new reference frame is less than $c$. When we switch reference frames, we use the velocity addition formula, which in this case is:
$$v' = \frac{2v}{1+v^2/c^2}$$
Note that it's a mistake to try to use this formula for separation velocity addition within a single reference frame.

The point about that formula is that if $v < c$, then $v' < c$.

 

[YouAreAwesome]

If I've got an object A going at 0.5c in one direction and an object B going at 0.5c in the opposite direction then the distance between them as measured by me grows at c. That's what Morin is talking about. That does not mean that A is moving away from B at c as measured by B. That's what you are talking about and you are correct that this is always less than c.

Thanks, this is at the core of my confusion. If B is traveling left at 0.5c, then B would measure me as traveling to the right at 0.5c, yes? But if I'm traveling at 0.5c to the right relative to B, what speed is A traveling relative to B? I would have thought A is traveling to the right at c. And then if we change 0.5 to 0.6 in this scenario, we break the speed of light. I can't see how this formula from @PeroK applies to the situation,

$v' = \frac{2v}{1+v^2/c^2}$

Does this calculate the speed of A relative to B?

 

[robphy]

This spacetime diagram on rotated graph paper might help.

I will use more arithmetically-convenient numbers.

In the lab frame,
the green observer moves to the left at (6/10)c, and
the blue observer moves to the right at (6/10)c.
And these observers separate with separation rate at (12/10)c=(1.2 light-years/yr) in the lab frame,
the rate (according to the lab) of relative-displacement of blue observer from the green observer per unit lab time.

But in the green frame,
the blue observer moves to the right with relative-velocity at (15/17)c=(0.8824 light-years/yr),
the slope of the blue worldline according to the green observer (which must be less than the speed of light).
The spacelike displacement PA is purely spatial according to the green observer, along OP.
[That is, P (the 17th tick according to the green observer) and the distant event A on the blue worldline are simultaneous according to the green observer.

$\begin{align*} v_{AB} &= \frac{v_{AL} - v_{BL}}{1- v_{AL} v_{BL}/c^2} \\ &= \frac{(6/10)c - (-6/10)c}{1- (6/10)(-6/10)} \\ &= \frac{15}{17}c \\ \end{align*}$

 

[Nugatory]

Does this calculate the speed of A relative to B?

If A is moving at speed $u$ relative to you, and B is moving at speed $v$ relative to A, then the the speed of B relative to you will be $(u+v)/(1+uv)$ (measuring distances in light-seconds and time in seconds so that $c=1$ and we don’t clutter up our equations with unnecessary and annoying factors of $c$ and $c^2$).

If instead you have the speed $u$ of A relative to you and the speed $v$ of B relative to you, you can restate the problem so that you are moving at speed $-u$ relative to A and B is moving relative to you.

 

[PeroK]

Thanks, this is at the core of my confusion. If B is traveling left at 0.5c, then B would measure me as traveling to the right at 0.5c, yes? But if I'm traveling at 0.5c to the right relative to B, what speed is A traveling relative to B? I would have thought A is traveling to the right at c. And then if we change 0.5 to 0.6 in this scenario, we break the speed of light. I can't see how this formula from @PeroK applies to the situation,

$v' = \frac{2v}{1+v^2/c^2}$

Does this calculate the speed of A relative to B?

Yes, for the special case of two objects traveling at the same speed in opposite directions. The more general "velocity addition" formula (for motion in 1D) is:
$$u' =\frac{u + V}{1 + uV/c^2}$$
Where $u$ is the velocity of a particle in some reference frame and $u'$ is its velocity in a different reference frame, moving at velocity $V$ relative to the first. I used this in the special case where $u = V = v$. Note that, in any case, if $u, V < c$, then $u' < c$.

You can also take $u = c$ and see how the velocity of light transforms to a new reference frame:
$$u' = \frac{c + V}{1 + V/c} = c$$
That formula, therefore, respects the invariance of the speed of light in any (inertial) reference frame! If something is moving at $c$ in one reference frame, it is moving at $c$ in any reference frame.

 

[jbriggs444]

Thanks, this is at the core of my confusion. If B is traveling left at 0.5c, then B would measure me as traveling to the right at 0.5c, yes? But if I'm traveling at 0.5c to the right relative to B, what speed is A traveling relative to B? I would have thought A is traveling to the right at c. And then if we change 0.5 to 0.6 in this scenario, we break the speed of light.

The problem with this naive formulation is that it combines velocities from two different frames as if they were taken from the same frame.

We have "me" traveling to the right at 0.5c relative to B. And we have A traveling to the right at 0.5c relative to "me". We want a result expressed as a velocity relative to B. Those input velocities are drawn from two different reference frames. They do not add directly. It would be a bit like adding apples and oranges.

By contrast if you take B traveling to the left at 0.5c relative to me and A traveling to the right at 0.5c relative to me then those add directly and yield a result relative to me. Everything in one frame. However, as elaborated previously, the result is not a velocity of anyone thing relative to another. It is a "separation rate" -- the rate at which a difference in positions is changing over time.

 

[PeroK]

... part of the confusion, of course, is that in Newtonian physics the two things are the same. If particles $A$ and $B$ have velocities $u$ and $v$ in one frame, then:

a) The relative velocity/separation velocity of $B$ relative to $A$ in that frame is $v - u$.

b) The velocity of $B$ in the rest frame of $A$ is also $v - u$.

In SR these two things are not the same. In SR, the calculation for b) must use the relativistic velocity addition formula given above.

 

[robphy]

A useful analogy is to compute relative slopes in Euclidean geometry.
Slopes don't add. (But angles do... that's why rapidity is useful.)

 

[YouAreAwesome]

The problem with this naive formulation is that it combines velocities from two different frames as if they were taken from the same frame.

We have "me" traveling to the right at 0.5c relative to B. And we have A traveling to the right at 0.5c relative to "me". We want a result expressed as a velocity relative to B. Those input velocities are drawn from two different reference frames. They do not add directly. It would be a bit like adding apples and oranges.

By contrast if you take B traveling to the left at 0.5c relative to me and A traveling to the right at 0.5c relative to me then those add directly and yield a result relative to me. Everything in one frame. However, as elaborated previously, the result is not a velocity of anyone thing relative to another. It is a "separation rate" -- the rate at which a difference in positions is changing over time.

I think this has been my difficulty. It is difficult for me to grasp how the velocity of A traveling away from me is not a concrete distance per time in all frames of reference. It seems instead, it varies according to who is measuring it. Thus A is traveling away from B at a different velocity to the velocity A is traveling away from me. B seems to be traveling at a different distance per time depending on who measures it. And this is weird. So A and I observe different things. Can we then say that both B and "me" are both correct in our observations? Because this means there is no such thing as objective velocity, but only velocity with respect to an observer. Weird.

 

[robphy]

reformatted by me

I think this has been my difficulty.
It is difficult for me to grasp how the velocity of A traveling away from me is not a concrete distance per time in all frames of reference.
It seems instead, it varies according to who is measuring it.
Thus A is traveling away from B at a different velocity to the velocity A is traveling away from me.
B seems to be traveling at a different distance per time depending on who measures it.
And this is weird.
So A and I observe different things.
Can we then say that both B and "me" are both correct in our observations?
Because this means there is no such thing as objective velocity, but only velocity with respect to an observer. Weird.

Try this. I've replaced "velocity" by signed-slope, etc...

I think this has been my difficulty.
It is difficult for me to grasp how the signed-slope of line A is not a fixed perpendicular-displacement per radial distance in all orientations of axes.
It seems instead, it varies according to which radial line is used as a reference.
Thus line A is angled away from line B at a different signed-slope to the signed-slope of line A angled away from line I.
Line B seems to be angled at a perpendicular-displacement per radial distance depending on which radial line is used as a reference.
And this is weird.
So line A and line I observe different signed-slopes.
Can we then say that both line B and line I are both valid in measuring signed-slopes ?
Because this means there is no such thing as fixed signed-slope , but only signed-slope with respect to a given line .
Weird.

 

[Ibix]

It is difficult for me to grasp how the velocity of A traveling away from me is not a concrete distance per time in all frames of reference.

Remember length contraction, time dilation, and the relativity of simultaneity. A's clocks and rulers aren't measuring the same thing yours are. Is it any surprise that the velocities A measures aren't generally the same as the ones you measure?

The underlying reason for this is the geometrical point @robphy is making. Loosely speaking, spacetime is a thing, and you and A are choosing different ways to split it up into space and time. You genuinely measure different distances and times because you are calling different slices through spacetime "space" and different directions "time".

 

[Mister T]

It is difficult for me to grasp how the velocity of A traveling away from me is not a concrete distance per time in all frames of reference.

The fact that we think it ought to be concrete seems embedded in our common sense, until we start to why.

The length of something is established by noting where its end points are at the same time. And it turns out that the notion of "at the same time" varies.

It seems instead, it varies according to who is measuring it. Thus A is traveling away from B at a different velocity to the velocity A is traveling away from me.

Well A moves away from B with a speed that differs from the speed A moves away from you, even when we use the Newtonian approximation.

 

[YouAreAwesome]

You can also take a look at the YouTube video (search for Paul Hewitt, Twin Trip). The scenario shown there is very close to what you describe it's just that light pules are used instead of light colors, but the same logic applies to both.

I love this video but I don't understand how the spaceship can be returning to Earth with the light pulses headed toward it so that when they collide the light is still traveling at the same speed as if the spaceship was at rest. I'm sorry I'm just not understanding how the spaceships velocity with regard to the light coming towards it is not a factor in measuring the speed of the light. Because the way we measure velocity is distance covered divided by time taken. Yet in the example where the spaceship heads towards the light pulse, and we take the spaceship as our frame of reference, then we can measure the light as traveling at a greater speed than c. I'm sure this has been explained already, and I've missed it over and over, but it's just not clicking yet... what's wrong with my brain :( I thought I was somewhat intelligent but this is making me question myself...

 

[jbriggs444]

Because the way we measure velocity is distance covered divided by time taken.

Time taken is arrival time minus departure time. Those are at different places. The relativity of simultaneity kicks in.

 

[Ibix]

Because the way we measure velocity is distance covered divided by time taken.

Yes - and the ship's rulers are length contracted and its clocks are time dilated and out of sync, according to the Earth frame. All of those effects combine to result in the measurement of $c$ being the same.

we take the spaceship as our frame of reference,

Pedantry, but important pedantry: you need to say "we take a frame of reference in which the spaceship is at rest", or else "we take the rest frame of the spaceship".

 

[PeroK]

I love this video but I don't understand how the spaceship can be returning to Earth with the light pulses headed toward it so that when they collide the light is still traveling at the same speed as if the spaceship was at rest. I'm sorry I'm just not understanding how the spaceships velocity with regard to the light coming towards it is not a factor in measuring the speed of the light. Because the way we measure velocity is distance covered divided by time taken. Yet in the example where the spaceship heads towards the light pulse, and we take the spaceship as our frame of reference, then we can measure the light as traveling at a greater speed than c. I'm sure this has been explained already, and I've missed it over and over, but it's just not clicking yet... what's wrong with my brain :( I thought I was somewhat intelligent but this is making me question myself...

Everything in physics is based on assumptions of some sort. Let's take the length of something. It was assumed in Newtonian physics that something has a definite length, independent of whether it is measured when at rest or while moving. That seems so logical that some people may say that their brain is not capable of comprehending any other possibility. And that may be true. It may be that many people could never learn SR because their brains have become hard-coded with the rules of Newtonian physics.

But, nevertheless, if you have a flexible mind you can analyse that assumption and ask if there is an alternative? And, in fact, if you start right back at the foundational assumptions about space and time you find there are two possibilities (not only one). Broadly, these possibilities are:

1) The absolute space and time of Newtonian physics.

2) The Einstein/Minkowski spacetime of SR.

Now you must go out and conduct an experiment to see which universe we actually live in. And, experiments show that it is the second of these: Minkowski spacetime. The initial assumption that space and time must be absolute is false. There is another possibility and experiment shows that that is the universe we live in. And if you do want to learn SR, then you must somehow learn to expand your thinking.

 

[Mister T]

Yet in the example where the spaceship heads towards the light pulse, and we take the spaceship as our frame of reference, then we can measure the light as traveling at a greater speed than c.

When you say take the spaceship as our frame of reference do you mean take the spaceship as being at rest in our frame of reference? There's no experiment that can be done within the spaceship that demonstrates it's in motion. You can look out the spaceship's window at Earth and see that Earth is approaching, but that simply means that Earth is moving in the spaceship's rest frame.

The effect of this relative motion between Earth and the spaceship is that people aboard the spaceship will observe the flashes arriving more frequently than people on Earth will observe them being sent, but both will agree on the speed of the light flashes.

 

[YouAreAwesome]

When you say take the spaceship as our frame of reference do you mean take the spaceship as being at rest in our frame of reference? There's no experiment that can be done within the spaceship that demonstrates it's in motion. You can look out the spaceship's window at Earth and see that Earth is approaching, but that simply means that Earth is moving in the spaceship's rest frame.

The effect of this relative motion between Earth and the spaceship is that people aboard the spaceship will observe the flashes arriving more frequently than people on Earth will observe them being sent, but both will agree on the speed of the light flashes.

Yes, that's what I am saying.

If I am watching a car travel towards me and I run towards it our combined velocity will be equal to the velocity I experience from my frame of reference. In the same way, let's say the spaceship is at rest and the light from Earth travels towards the spaceship at c. The light is equivalent to the car in this analogy and the spaceship is equivalent to me starting at rest. But before the light reaches the spaceship we decide to travel towards the light which reduces the distance the light has to travel, just as I run towards the car. However, from within the spaceship we do not start approaching the light with more velocity or else it would appear from within the spaceship that light is traveling faster than c (that is, c plus the the speed of the spaceship). My movement in the spaceship towards the light does not add or subtract velocity from the speed of the light in the way my running towards a car increases the combined velocity. What I am learning is that an outside observer will see the light traveling at c towards the spaceship, and they would also see the spaceship traveling towards the light. But from inside the spaceship the light would still appear to be traveling at the same speed, it would not appear to be traveling at the combined velocity of c plus the spaceship velocity. The fact the light hits the spaceship more quickly is the strange part. And I'm thinking the answer is because my speed takes me into a different slope of spacetime. And that's the situation I'm still pondering on. I've read this thread over and over and I think I am getting somewhere, but am I?

 

[PeroK]

Yes, that's what I am saying.

If I am watching a car travel towards me and I run towards it our combined velocity will be equal to the velocity I experience from my frame of reference. In the same way, let's say the spaceship is at rest and the light from Earth travels towards the spaceship at c. The light is equivalent to the car in this analogy and the spaceship is equivalent to me starting at rest. But before the light reaches the spaceship we decide to travel towards the light which reduces the distance the light has to travel, just as I run towards the car. However, from within the spaceship we do not start approaching the light with more velocity or else it would appear from within the spaceship that light is traveling faster than c (that is, c plus the the speed of the spaceship). My movement in the spaceship towards the light does not add or subtract velocity from the speed of the light in the way my running towards a car increases the combined velocity. What I am learning is that an outside observer will see the light traveling at c towards the spaceship, and they would also see the spaceship traveling towards the light. But from inside the spaceship the light would still appear to be traveling at the same speed, it would not appear to be traveling at the combined velocity of c plus the spaceship velocity. The fact the light hits the spaceship more quickly is the strange part. And I'm thinking the answer is because my speed takes me into a different slope of spacetime. And that's the situation I'm still pondering on. I've read this thread over and over and I think I am getting somewhere, but am I?

You're getting there. Put simply, we have either a) the invariance of the speed of light; or, b) absolute space and time. But not both. If we experimentally measure the invariance of the speed of light, then something's got to give in terms of our classical notion of absolute space and time. And one consequence of this is that we no longer have the classical velocity addition formula. Instead, we have the relativistic formula quoted in a previous post.

 

[Ibix]

If I am watching a car travel towards me and I run towards it our combined velocity will be equal to the velocity I experience from my frame of reference.

Not quite. You and the car obey the same velocity addition rule as everything else$$u'=\frac{u+v}{1+uv/c^2}$$But if you run at $u=20\mathrm{mph}$ (a decent sprint) and the car is doing $v=60\mathrm{mph}$, and we note that $c=6.75×10^8\mathrm{mph}$ then you can calculate that $uv/c^2=2.6×10^{-15}$. That means that the speed of the car in your reference frame is $(20+60)/(1+2.6×10^{-15})\mathrm{mph}$, which differs from 80mph by about the diameter of an atom per hour. So you can be forgiven for not noticing! But as the velocities get higher, $uv/c^2$ gets larger, and you will start to notice problems if you just add velocities.

This is generally true - all relativistic formulae produce answers very close to the Newtonian equivalent as long as you are moving a lot slower than $c$. But if you try to use the Newtonian formula at high speed (such as adding $v$ and $c$) you'll get wrong answers.

 

[YouAreAwesome]

Not quite. You and the car obey the same velocity addition rule as everything else$$u'=\frac{u+v}{1+uv/c^2}$$But if you run at $u=20\mathrm{mph}$ (a decent sprint) and the car is doing $v=60\mathrm{mph}$, and we note that $c=6.75×10^8\mathrm{mph}$ then you can calculate that $uv/c^2=2.6×10^{-15}$. That means that the speed of the car in your reference frame is $(20+60)/(1+2.6×10^{-15})\mathrm{mph}$, which differs from 80mph by about the diameter of an atom per hour. So you can be forgiven for not noticing! But as the velocities get higher, $uv/c^2$ gets larger, and you will start to notice problems if you just add velocities.

This is generally true - all relativistic formulae produce answers very close to the Newtonian equivalent as long as you are moving a lot slower than $c$. But if you try to use the Newtonian formula at high speed (such as adding $v$ and $c$) you'll get wrong answers.

Thanks so much for the replies. I have another way of explaining this problem that will hopefully help me iron out some more wrinkles.

Imagine I am superman (shouldn't be that difficult). A photon of light is headed towards me at velocity c. I am at rest, waiting to see the photon appear. I get impatient and I decide to run towards the photon so as to see it sooner. Lucky for me I can run at close to the speed of light. But as I run, the photon continues to come towards me at the same velocity. So what changes by me running? Well the distance between the photon and I reduces, but if distance reduces over some time then that velocity should be added to c, however it can not be. Therefore something else is changing. So is the thing changing spacetime? As in, as I run faster, because I move through a changing "slope" (axis?) of spacetime, do I see the photon sooner because of this change and not because the photon is moving towards me faster?

Because I keep thinking, just because I run towards something doesn't mean the thing begins moving faster. But relative to my rest frame it does move faster. However with light, relative to my rest frame, whether I am moving towards it or not, it doesn't move faster. But rather the distance and time I experience changes so it will appear to me that the light doesn't need to travel as far and will do so over a shorter time. Hmmm.

 

[PeroK]

Thanks so much for the replies. I have another way of explaining this problem that will hopefully help me iron out some more wrinkles.

Imagine I am superman (shouldn't be that difficult). A photon of light is headed towards me at velocity c. I am at rest, waiting to see the photon appear. I get impatient and I decide to run towards the photon so as to see it sooner. Lucky for me I can run at close to the speed of light. But as I run, the photon continues to come towards me at the same velocity. So what changes by me running? Well the distance between the photon and I reduces, but if distance reduces over some time then that velocity should be added to c, however it can not be. Therefore something else is changing. So is the thing changing spacetime? As in, as I run faster, because I move through a changing "slope" (axis?) of spacetime, do I see the photon sooner because of this change and not because the photon is moving towards me faster?

Because I keep thinking, just because I run towards something doesn't mean the thing begins moving faster. But relative to my rest frame it does move faster. However with light, relative to my rest frame, whether I am moving towards it or not, it doesn't move faster. But rather the distance and time I experience changes so it will appear to me that the light doesn't need to travel as far and will do so over a shorter time. Hmmm.

I would say that this is unsuccessfully trying to replace calculations with words. You need a formal analysis of your scenario and solid calculations.

Ultimately, most of your analysis is not precise enough to have any physical meaning.

 

[jbriggs444]

So what changes by me running?

Your frame of reference.

There is a mental picture that helped me get a handle on "frame of reference" in relativity. There is a three dimensional array of clocks all keeping pace with me in the middle. All of these clocks all run at the same rate. And they are all synchronized. They strike noon at the same time. The clocks are laid out on a set of cartesian coordinates, one clock per unit distance.

If I want to know the coordinate of some event in space-time, I find the clock nearest the event and read off its (x, y, z) coordinate. And I fill in t from the time that clock read when the event occurred.

"What changes by me running"...

I have to get all of my clocks accelerated to my new speed, laid out on my new length-contracted grid and synchronized with my new standard of synchronization.

 

[YouAreAwesome]

Your frame of reference.

There is a mental picture that helped me get a handle on "frame of reference" in relativity. There is a three dimensional array of clocks all keeping pace with me in the middle. All of these clocks all run at the same rate. And they are all synchronized. They strike noon at the same time. The clocks are laid out on a set of cartesian coordinates, one clock per unit distance.

If I want to know the coordinate of some event in space-time, I find the clock nearest the event and read off its (x, y, z) coordinate. And I fill in t from the time that clock read when the event occurred.

"What changes by me running"...

I have to get all of my clocks accelerated to my new speed, laid out on my new length-contracted grid and synchronized with my new standard of synchronization.

Thanks for that! So comparing my initial reference frame, where I was standing still, to the frame of reference where I am moving, I need to do the calculations for time and distance transformations as these change according to the velocity I am traveling? As in, time and length are literally different due to the shift in spacetime? So in my own "rest" frame, no matter if I am still or running, it appears that nothing has changed with regard to time and space. But if we compared the two, time and distance has indeed changed because I've rotated through the "spacetime" plane. Please tell me I get it... because I feel like I am starting to grasp the twins paradox... but I don't want to think I get it, when I don't.

 

[jbriggs444]

Thanks for that! So comparing my initial reference frame, where I was standing still, to the frame of reference where I am moving, I need to do the calculations for time and distance transformations as these change according to the velocity I am traveling? As in, time and length are literally different due to the shift in spacetime? So in my own "rest" frame, no matter if I am still or running, it appears that nothing has changed with regard to time and space. But if we compared the two, time and distance has indeed changed because I've rotated through the "spacetime" plane. Please tell me I get it... because I feel like I am starting to grasp the twins paradox... but I don't want to think I get it, when I don't.

This sounds very much like you are starting to "get it".

It is not that space-time has changed really. It is that the coordinate system you've laid down over it has changed.

Still, it is not just a trick of coordinates. The underlying space-time geometry is such that all of these coordinate systems actually work. And work identically.

 

[Mister T]

If I am watching a car travel towards me and I run towards it our combined velocity will be equal to the velocity I experience from my frame of reference.

Yes, but the way you combine them is not by adding them. This is analogous to stacking a pair of wedges. You set the first wedge on a level table top and measure the angle that the upper surface of the wedge makes with the table top. You also measure the slope of that upper surface. Now you stack a second wedge on top of the first wedge in such a way that the upper surface of the second wedge forms a steeper slope. You can add the angles of the wedges to get the angle of the upper surface. But you cannot add the slopes to get the slope of the upper surface. However, if the slopes are small adding them gives a good approximation. Likewise, adding speeds gives a good approximation when the speeds are slow.

My movement in the spaceship towards the light does not add or subtract velocity from the speed of the light in the way my running towards a car increases the combined velocity.

We would use the same formula in both cases.

But from inside the spaceship the light would still appear to be traveling at the same speed, it would not appear to be traveling at the combined velocity of c plus the spaceship velocity.

Yes it would, provided that you combine them properly. Adding them is not the correct way to combine them.

The fact the light hits the spaceship more quickly is the strange part.

But it doesn't hit it more quickly. The flashes of light arrive more frequently because they are closer together.

If you have a continuous beam of light (instead of flashes) then you note that the crests of the light wave arrive more frequently because they are closer together, not because they move at a higher speed.

 

[YouAreAwesome]

Yes, but the way you combine them is not by adding them. This is analogous to stacking a pair of wedges. You set the first wedge on a level table top and measure the angle that the upper surface of the wedge makes with the table top. You also measure the slope of that upper surface. Now you stack a second wedge on top of the first wedge in such a way that the upper surface of the second wedge forms a steeper slope. You can add the angles of the wedges to get the angle of the upper surface. But you cannot add the slopes to get the slope of the upper surface. However, if the slopes are small adding them gives a good approximation. Likewise, adding speeds gives a good approximation when the speeds are slow.
View attachment 264738
We would use the same formula in both cases.
Yes it would, provided that you combine them properly. Adding them is not the correct way to combine them.
But it doesn't hit it more quickly. The flashes of light arrive more frequently because they are closer together.

If you have a continuous beam of light (instead of flashes) then you note that the crests of the light wave arrive more frequently because they are closer together, not because they move at a higher speed.

With the triangles you have shown, where is distance, time and velocity? What do the angles in the triangles represent? Is tan of the angle the velocity, the opposite side the distance and the adjacent side time?

 

[jbriggs444]

With the triangles you have shown, where is distance, time and velocity? What do the angles in the triangles represent? Is tan of the angle the velocity, the opposite side the distance and the adjacent side time?

It is a Euclidean analogy to a Minkowsky geometry.

An acceleration in Minkowski space corresponds to a rotation in Euclidean space.

The origin for the drawing it the common vertex on the right.

You have the original trajectory (the lower line) which is a time elapsed horizontally (horizontal = time) and a certain distance vertically (vertical = space).

You have the the first change in velocity. (the lower triangle). You have an additional distance moved (the vertical line on the left). This takes place over the original horizontal time line. This yields a new trajectory (the diagonal line on the lower triangle).

Now you reset your frame of reference so that this diagonal line is your new baseline trajectory. It is your new horizontal time line. Its "horizontal" length is elapsed time. The "vertical" distance moved is zero. This new baseline is at zero velocity.

The next short segment on the left is a displacement in this new frame. Over the duration of the trajectory some object moves that much distance. Its trajectory is the long upper diagonal line in the upper triangle. You can read this trajectory as distance in the original frame versus time in the original frame. Or you can rotate your point of view and read it as distance in the second frame versus time in the second frame.

 

[YouAreAwesome]

This sounds very much like you are starting to "get it".

It is not that space-time has changed really. It is that the coordinate system you've laid down over it has changed.

Still, it is not just a trick of coordinates. The underlying space-time geometry is such that all of these coordinate systems actually work. And work identically.

Awesome, thanks.

So back to the original idea that a spaceship is traveling back to Earth in view of a video projection from Earth of a clock, does the spaceship see the clock tick faster on the return journey? If yes, is it because the spacetime coordinate system "rotates"? I'm thinking it does tick faster, and it's quite a scary thought in practise. If one day we can travel at close to light speed, and we find ourselves far, far away from earth, then the only way back would be to arrive at an Earth that has aged many years. Thus we don't return to our loved ones. The faster we return, the less time we get to spend with them. A sad thought.