Effects of time dilation for near-speed-of-light travel

[PeterDonis]

My analysis uses a non accelerated reference frame during the first half, and then an accelerated reference frame during the second half

I'm not sure which analysis of yours you are referring to here. I don't see this in any of your posts in this thread.

You should also be aware that constructing a valid accelerated reference frame is a non-trivial task. You should not attempt it until you have a thorough grasp of inertial frames. Nor do you need to to analyze the scenarios under discussion; they can all be analyzed just fine using inertial frames, and since the key quantities are invariants, it doesn't matter what frame you use to calculate them.

 

[student34]

I did say that there was some asymmetric (it's possible that was a bad choice of word on my part) relationship between what the two observers saw, statement 1 = clock A is younger and clock B is older, or statement 2 = clock A is older and clock B is younger; if I'm not mistaken, statement 1 and statement 2 cannot both be true simultaneously for proper time measurements, but one of them can be true or neither can be true for proper time measurements.

You are missing a very crucial concept of time dilation. Time dilation does not actually happen to anyone/anything. It is only something that happens when comparing clocks to other clocks. So the clock in the spaceship is only running slower in comparison to a clock on Earth.

 

[FactChecker]

You are missing a very crucial concept of time dilation. Time dilation does not actually happen to anyone/anything. It is only something that happens when comparing clocks to other clocks. So the clock in the spaceship is only running slower in comparison to a clock on Earth.

Although in some situations, like the Twin Paradox, the scenario will bring them together where they are both using the same clock. In those cases, the effects of the intermediate slower clocks and processes are self-evident.

 

[Sagittarius A-Star]

They are numerically equal if you use an inertial frame

More precisely, that's exactly valid only momentarily in a co-moving inertial reference frame. The more general formula is $\alpha = \gamma^3 a$, with proper acceleration $\alpha$ and coordinate acceleration $a$.

Source:
https://en.wikipedia.org/wiki/Proper_acceleration#Acceleration_in_(1+1)D

 

[Chenkel]

@Chenkel, please read this:

https://www.physicsforums.com/threads/when-discussing-the-twin-paradox-read-this-first.1048697/

I'm reading the article, currently I am here.

The author writes:$$d\tau = dt^2 - dx^2 - dy^3 - dz^2$$One question I have about this is how can you subtract velocity squared units from time squared units?

I'm not sure what I'm missing here.

Also I was studying Lorentz transformations (not sure if this is related to the above, but I wanted to brush up for the article.) and the definition of the Lorentz transformations here.

It seems to me based on what I read I can write the following using the "Lorentz factor":$$t' = \gamma(t - \frac {vx} {c^2})$$$$x' = \gamma(x - vt)$$One thing I was wondering is if $x'$ is considered the stationary reference frame (at least this seems to be the case to me, but I'm not sure) based on the wikipedia page entry, then I was wondering if I can change the transform so that $x$ is considered the stationary reference frame by simply doing the following transformation to velocity $v := -v$.

If the transformation is symetric around velocity, does that mean that the two reference frames will agree on the traveled distance regardless of which reference frame we choose as the "stationary" one?

Hopefully my intuition is not deceiving me.

I appreciate any feedback on these concepts, thanks again for the article!

 

[PeterDonis]

how can you subtract velocity squared units from time squared units?

You're not. The author is using units in which $c = 1$, so all of the quantities are in units of either time squared or distance squared, however you want to think of it.

It seems to me based on what I read I can write the following using the "Lorentz factor":$$t' = \gamma(t - \frac {vx} {c^2})$$$$x' = \gamma(x - vt)$$

Yes. Note that in relativity it's common to use units in which $c = 1$. Notice that the two equations look more symmetric that way.

One thing I was wondering is if $x'$ is considered the stationary reference frame

Usually that's the frame without primes, but of course "stationary" is relative, you can look at it either way. There is no such thing as absolute rest in relativity.

I was wondering if I can change the transform so that $x$ is considered the stationary reference frame by simply doing the following transformation to velocity $v := -v$.

Yes, that's one way of obtaining the inverse of a Lorentz transformation.

If the transformation is symetric around velocity, does that mean that the two reference frames will agree on the traveled distance regardless of which reference frame we choose as the "stationary" one?

No. The two frames will agree on spacetime intervals between specific events, but "traveled distance" is not such a thing. You can express "traveled distance" in a specific frame as a spacetime interval between two events, but which two events is frame-dependent; in a different frame it will be a different two events, so a different interval.

 

[PeterDonis]

The author writes:$$d\tau = dt^2 - dx^2 - dy^3 - dz^2$$

Not quite. The LHS is $d\tau^2$, not $d\tau$. Also, on the RHS it is $dy^2$, not $dy^3$.

 

[Chenkel]

Not quite. The LHS is $d\tau^2$, not $d\tau$. Also, on the RHS it is $dy^2$, not $dy^3$.

My bad, perhaps this is better?$$d\tau^2= dt^2 - dx^2 - dy^2 - dz^2$$I'm not sure how we get units of time on the LHS, when we are subtracting units of (L^2/T^2) from units of T^2 on the RHS.

 

[Chenkel]

My bad, perhaps this is better?$$d\tau^2= dt^2 - dx^2 - dy^2 - dz^2$$I'm not sure how we get units of time on the LHS, when we are subtracting units of (L^2/T^2) from units of T^2 on the RHS.

You're not. The author is using units in which c=1, so all of the quantities are in units of either time squared or distance squared, however you want to think of it.

Just read this, thanks for letting me know, I'll see if I can understand the article in that context.

 

[PeterDonis]

My bad, perhaps this is better?$$d\tau^2= dt^2 - dx^2 - dy^2 - dz^2$$

Yes.

I'm not sure how we get units of time on the LHS

You don't. It's units of time squared (or distance squared, your choice--see below). The formula is a formula for a squared interval.

when we are subtracting units of (L^2/T^2) from units of T^2 on the RHS.

No, you're not. All of the quantities on the RHS have units of either L^2 or T^2, it's your choice which, since the formula is written in units where $c = 1$, so the units of L and T are the same. There is no L^2 / T^2 anywhere.

 

[Chenkel]

No. The two frames will agree on spacetime intervals between specific events, but "traveled distance" is not such a thing. You can express "traveled distance" in a specific frame as a spacetime interval between two events, but which two events is frame-dependent; in a different frame it will be a different two events, so a different interval.

Perhaps I should stop thinking of velocity in terms of meters per second, where velocity equals (traveled distance) / seconds, and instead start thinking of velocity in terms of meters per light second (a unit of time) instead of as meters per second (a rate of change that might be reference frame dependent), because all velocities should be thought as relative to the speed of light.

 

[PeterDonis]

all velocities should be thought as relative to the speed of light

In relativity with units where $c = 1$, yes, velocities are dimensionless numbers with a maximum of $1$ (the speed of light).

 

[PeterDonis]

we are subtracting units of (L^2/T^2)

Why do you think there are units of L^2 / T^2 on the RHS? The interval formula has no velocities anywhere.

 

[DrGreg]

My bad, perhaps this is better?$$d\tau^2= dt^2 - dx^2 - dy^2 - dz^2$$

It is perhaps worth commenting that if you don't like using units where $c = 1$, the formula becomes
$$c^2 \, d\tau^2= c^2 \, dt^2 - dx^2 - dy^2 - dz^2$$

 

[Chenkel]

Why do you think there are units of L^2 / T^2 on the RHS? The interval formula has no velocities anywhere.

I'm not sure, my intuition was telling me that it contained velocities (my intuition is saying something else now); I'm still trying to make sense of the diagram, and the formula in the article, and how proper time works.

I initially thought dx, dy, and dz were velocities because I confused dx, dy, and dz, with dx/dt, dy/dt, and dz/dt, I need to read the article again to see exactly what dx, dy, and dz represent, currently I'm guessing dx is some kind of differential representing an amount of time or length.

I'm starting to think that a good way to treat velocity is meters per light second, I accidentally said in the previous post that the dimensions of this would be in seconds, now I am guessing that it would actually be dimensionless based on my possible understanding of how the units cancel.

The reasoning I have behind velocity being dimensionless is that velocity should be meters per light second and this should theoretically be dimensionless based on the following analysis velocity = (meters per second) * (seconds per light second) = (L/T)*(T/L) so the units cancel.

I reason c=1 because c = (299792458 meters per second) * (1 second / 299792458 meters)) = (meters per second) * (seconds per light second) = (L/T)*(T/L).

If we say c is always one, can we still somehow say c = 299792458 meters / sec, as some people like to do? What happens to the nomenclature for the constant 299792458 meters per second when we define c as dimensionless one?

Thanks again for all the help!

 

[PeterDonis]

I'm guessing dx is some kind of differential representing an amount of time or length

All of the terms on the RHS are squares of coordinate differentials. The idea is that you are evaluating arc length along a curve; $d\tau$ is the arc length along a small differential element of the curve. $dt$, $dx$, $dy$, and $dz$ are the coordinate differentials along that small differential element of the curve.

I'm starting to think that a good way to treat velocity is meters per light second

No, that's not a good way, because in those units $c$ is not 1. If you measure distance in meters and $c = 1$, then the unit of time is meters--a meter of time is the time it takes light to travel 1 meter (about 3.3 nanoseconds).

If you measure time in seconds, and $c = 1$, then distance is measured in light seconds (seconds of distance).

I reason c=1 because c = (299792458 meters per second) * (1 second / 299792458 meters)) = (meters per second) * (seconds per light second) = (L/T)*(T/L).

This is not valid. See above.

If we say c is always one, can we still somehow say c = 299792458 meters / sec, as some people like to do?

No. See above.

What happens to the nomenclature for the constant 299792458 meters per second

There is no such constant in units where $c = 1$. That constant is part of a specific system of units, SI units.

 

[Chenkel]

All of the terms on the RHS are squares of coordinate differentials. The idea is that you are evaluating arc length along a curve; $d\tau$ is the arc length along a small differential element of the curve. $dt$, $dx$, $dy$, and $dz$ are the coordinate differentials along that small differential element of the curve.No, that's not a good way, because in those units $c$ is not 1. If you measure distance in meters and $c = 1$, then the unit of time is meters--a meter of time is the time it takes light to travel 1 meter (about 3.3 nanoseconds).

If you measure time in seconds, and $c = 1$, then distance is measured in light seconds (seconds of distance).This is not valid. See above.No. See above.There is no such constant in units where $c = 1$. That constant is part of a specific system of units, SI units.

I am a little confused on the units aspect of relativity.

Why do some people write c = 1?

Why do some people write c = 299792458 meters per second?

I would rather just write c = 1 if I can get away with it, it seems it can potentially simplify calculations and it seems like physicists are fond of doing it, and that is one of the reasons I am curious on how to be effective in this approach.

It's a puzzle to me because I'm wondering what units c = 1 means in the language of the problem that is being solved, if c is defined as the speed of light, then what units is c = 1, and why aren't dimensions specified?

How do people utilize an effective language for describing problems using these units constructs?

I've heard when someone sets the speed of light to 1, they are using "natural units."

Does the 1 in the expression "c = 1" represent one light second per second? And we just say one light second per second instead saying 299792458 meters per second? Is it common to not specify the dimensions of c explicitly, but to just refer to it as the speed of light?

 

[Ibix]

I personally had a bit of trouble with blithely setting $c=1$ (it makes more sense the further you get into the geometric picture of relativity). It helped me to explicitly write down such a unit system, and there's a really easy one: measuring distance in light seconds and time in seconds. The speed of light is, of course, one light second per second.

Light speed is also approximately one foot per nanosecond. Somebody (edit: David Mermin in his book "It's About Time", p22, according to Wikipedia) semi-jokingly proposed the phoot as a name for the light nanosecond...

 

[Chenkel]

I personally had a bit of trouble with blithely setting $c=1$ (it makes more sense the further you get into the geometric picture of relativity). It helped me to explicitly write down such a unit system, and there's a really easy one: measuring distance in light seconds and time in seconds. The speed of light is, of course, one light second per second.

Light speed is also approximately one foot per nanosecond. Somebody (don't recall who now) semi-jokingly proposed the phoot as a name for the light nanosecond...

You had me chuckling with the "phoot."

So when one writes c = 1, you are using "natural units" and are essentially implicitly implying you are measuring distance in light seconds, and time in seconds!

That seems to make some sense to me.

But if I asked someone "how fast was the photon going?" Even if we are using natural units I would want them to say "one light second per second" or "the speed of light" and not just respond with the singular word "one." If they want to write down v = c I will just read "ah yes, the photon is traveling at the speed of light, one light second per second!"

 

[malawi_glenn]

velocity squared units from time squared units?

you mean "time squared units"
and you forgot to square dτ

 

[malawi_glenn]

Why do some people write c = 1?

because algebraically, it does not matter what the value of $c$ is.
you can always insert the missing factors of $c$ in your final expression.
every intro book on relativity explains this.

If they want to write down v = c I will just read "ah yes, the photon is traveling at the speed of light, one light second per second!"

Nope you can not do that, because $v$ is the relative speed of the two interial frames in the lorentz-transformation, which is not valid for $v=c$.

 

[Chenkel]

because algebraically, it does not matter what the value of $c$ is.
you can always insert the missing factors of $c$ in your final expression.
every intro book on relativity explains this.Nope you can not do that, because $v$ is the relative speed of the two interial frames in the lorentz-transformation, which is not valid for $v=c$.

Doesn't the clock for a photon basically never tick for anyone observing the photons clock because the Lorentz factor increases without bound as velocity approaches c? What's wrong with plugging in v=c in the Lorentz factor and just treating the Lorentz factor as a infinite hyperreal?

 

[Chenkel]

because algebraically, it does not matter what the value of $c$ is.
you can always insert the missing factors of $c$ in your final expression.
every intro book on relativity explains this.Nope you can not do that, because $v$ is the relative speed of the two interial frames in the lorentz-transformation, which is not valid for $v=c$.

My last reply just talked about treating Lorentz factor as an infinite hyperreal, not sure if this can be done, but I'm sure my analysis is not on point considering my lack of knowledge in special relativity.

 

[Ibix]

So when one writes c = 1, you are using "natural units" and are essentially implicitly implying you are measuring distance in light seconds, and time in seconds!

Or years and light years (useful because $g$ is close to 1 light year/year² - actually 1.032), months and light months, whatever.

But if I asked someone "how fast was the photon going?" Even if we are using natural units I would want them to say "one light second per second" or "the speed of light" and not just respond with the singular word "one." If they want to write down v = c I will just read "ah yes, the photon is traveling at the speed of light, one light second per second!"

As you get more into the geometry it makes more sense that you want to use the actual same units for distance and time. The Lorentz transforms are the Minkowski geometry equivalent of rotating a Cartesian coordinate system in Euclidean geometry. In Euclidean geometry there's nothing wrong with measuring vertical distances in fathoms and horizontal ones in nautical miles, but it's easier if you use the same units for everything. Thus some textbooks will explicitly start measuring time in meters (one three hundred millionth of a second) or something.

Personally I set $c=1$ and don't worry about whether it's "actually" 1 or "actually" 1ls/s. The maths is the same and I can always insert the $c$s again by dimensional analysis if necessary.

 

[malawi_glenn]

I'm sure my analysis is not on point considering my lack of knowledge in special relativity.

Why not just pick up a book about relativity and study it? I can recommend the book by Morin "Special Relativity: For the Enthusiastic Beginner" which is quite cheap.

It is hard (impossible) to learn a new subject on a forum.

 

[Chenkel]

Why not just pick up a book about relativity and study it? I can recommend the book by Morin "Special Relativity: For the Enthusiastic Beginner" which is quite cheap.

Thanks for the reference, I'll check it out!

 

[jbriggs444]

Light speed is also approximately one foot per nanosecond.

Grace Hopper used to hand out nanoseconds at her lectures. (one foot wire segments). The one time I heard her speak, wire was too expensive and I picked up a salt packet of picoseconds instead. (Google says pepper. I remember salt. Go figure).

 

[robphy]

Grace Hopper used to hand out nanoseconds at her lectures. (one foot wire segments). The one time I heard her speak, wire was too expensive and I picked up a salt packet of picoseconds instead. (Google says pepper. I remember salt. Go figure).

Go to t=46m00s [for milliseconds... start at t=45m07s]



About Adm Hopper


https://en.wikipedia.org/wiki/Grace_Hopper

 

[PeterDonis]

Why do some people write c = 1?

Because they're using units called "natural units" in which $c = 1$.

Why do some people write c = 299792458 meters per second?

Because they're using SI units, in which ##c# is defined to have that value.

I'm not sure what the issue is with this. Surely the existence of multiple different systems of units is no mystery.

Does the 1 in the expression "c = 1" represent one light second per second?

It can. Or it can mean one meter per light-meter (meter of light travel time). Or it can mean one light year per year. The point is that the unit of distance and the unit of time are related by the time it takes light to travel the distance. That makes things much simpler mathematically (you don't have stray factors of $c$ all over the place and have to worry about whether you've gotten them all right) and makes spacetime diagrams easier (because the worldlines of light rays are 45 degree lines and the units on both the space and time axes are the same).

 

[PeterDonis]

My last reply just talked about treating Lorentz factor as an infinite hyperreal, not sure if this can be done

No, it can't. There is no such thing as an inertial frame in which a photon is at rest.

 

[Chenkel]

Why not just pick up a book about relativity and study it? I can recommend the book by Morin "Special Relativity: For the Enthusiastic Beginner" which is quite cheap.

It is hard (impossible) to learn a new subject on a forum.

I got the book recently and I've been studying it, it's interesting so far and I think I'll learn a lot from it.

Thank you for the suggestion