How Do the Axes of the Minkowski Diagram Work?

[JohnnyGui]

Hello,

First of all, I'm sorry since I bet there are quite a few threads about this but I still have a bit of a hard time wrapping the axes of the Minkowski diagram around my head.

I understand very well that, to have the speed of light traveling at a 45 degree angle in a space-time diagram, one could change the time (t) and distance (x) axes in 2 ways since the length of each t and each x unit are the same in the diagram:
1. Change the t units into time units in which light travels one x unit distance, thus the vertical t axis having units of x / c time. (since distance / velocity equals the time that the speed of light travels one x distance)
2. Change the x units into distance units that light travels in one unit time, thus the horizontal x-axis having units of c ⋅ t distance (since velocity times time gives the distance)

However, I can't for the life of me get how a time axis of c ⋅ t would give a 45 degree angle of the speed of light as well. I know that c ⋅ t will give a distance unit but that means that the space-time diagram will have 2 distance axes. How can light (or anything else for that matter) therefore travel at a distance unit per distance unit?? Looking at it this way would make me say that for each square in the diagram, light travels a 300.000.000m per meter which is hard for me to understand.

Is there a clear way to explain the meaning of 2 distance axes and, maybe formula wise, to explain why lightspeed would have a 45 degree angle in such a case? I'd appreciate the help a lot.

 

[PeterDonis]

I know that c ⋅ t will give a distance unit but that means that the space-time diagram will have 2 distance axes.

No, it doesn't. Using distance units on the time axis does not make time into distance. It just makes the units of time more convenient, since the speed of light is 1 in these units. Time is still physically different; timelike and spacelike intervals still have opposite signs in the Minkowski metric.

Looking at it this way would make me say that for each square in the diagram, light travels a 300.000.000m per meter

No, it doesn't. In the units you are talking about, light travels at speed 1, a unitless number. You can think of it as "1 meter per meter", where a "meter" of time equals 3.3 nanoseconds, the time in conventional units that it takes light to travel 1 meter of distance in conventional units. Or you can think of it as "1 second per second", where a "second" of distance is what we usually call a "light-second", i.e., the distance in conventional units that light travels in 1 second of time in conventional units, i.e., $3 \times 10^8$ meters.

To see how this comes out from using $ct$ instead of $t$, look at it this way: in conventional units, we have

$$
v = \frac{x}{t}
$$

Here $v$ has units of meters per second. In "natural" units, we divide both sides of this by $c$, to get

$$
\beta = \frac{v}{c} = \frac{x}{ct}
$$

Here $\beta$ is a unitless number, and if $v = c$, we can see that $\beta = 1$, so the speed of light is 1 in these units. For that case, if you set x = 1 meter, you must have ct = 1 meter, so t = 1 / c or 3.3 nanoseconds. If you set x = $3 \times 10^8$ meters, then you must have ct = $3 \times 10^8$ meters, so t = 1 second. And so on.

 

[Mister T]

Looking at it this way would make me say that for each square in the diagram, light travels a 300.000.000m per meter which is hard for me to understand.

If a grid square measured $3 \times 10^8$ meters on one side, it would measure $1$ second on the adjacent sides. Thus the 45° diagonal line you mentioned would have a slope of $3 \times 10^8$ meters per second.

It's natural to simply refer to a distance of $3 \times 10^8$ meters as a distance of $1$ s.

Or alternatively to refer to a time of $\frac{1}{3 \times 10^8}$ seconds as $1$ meter of time. One very nice reason why this is natural is the meter is actually defined as the distance light travels in $\frac{1}{299\ 792 \ 458}$ seconds.

In these natural units that line would have a slope of $1$.

 

[Ibix]

I find the easiest way to think about it is that the time unit is one second and the distance unit is one light second (or one year and one light year, etc.). You can always convert back to SI units by multiplying by the appropriate factor if you need the answer in those units.

 

[Dale]

I know that c ⋅ t will give a distance unit but that means that the space-time diagram will have 2 distance axes. How can light (or anything else for that matter) therefore travel at a distance unit per distance unit??

This concept is called geometrized units. Basically, time is just considered a length, like meters, but in a different direction. Then a spacetime diagram has units of length like a normal space only diagram. Just like in a normal diagram lines have slopes and the slope is a dimensionless number, similarly speed is a dimensionless number normalised to the speed of light.

 

[JohnnyGui]

Thanks a lot for the answers. Can I say that giving c a value of 1 without units, means that everything else that moves in the spacetime diagram is given a particular velocity with respect to c? So to know their true velocities with respect to 1 meter, one would have to multiply the slope of every moving object in the diagram with c to get its true velocity?

Wouldn't one achieve the same thing by giving the units in the time axis just a value of x/c instead of ct? Every moving object in that case will have to be multiplied by c as well (their velocities are with respect to c as well) to get its true velocity, since their slope would be:
x_{object traveld in x/c time} / (x / c) which gives a fraction of c which then must be multiplied by c to get the velocity of the object. I think in this case c would have a number with units but does it really matter to have a number 1 without units if you achieve the same thing with units?

 

[PeterDonis]

Can I say that giving c a value of 1 without units, means that everything else that moves in the spacetime diagram is given a particular velocity with respect to c?

This isn't exactly wrong as you state it, but it appears to be leading you to confuse yourself. You are making this much more difficult than it needs to be. Drawing a spacetime diagram in units where $c = 1$ is just a convenient choice of units; that's all it is. If it helps, think of it as picking units so $c$ is 1 light-second per second, or 1 light-year per year, or something like that. Then you can just think about the units of the $x$ axis in terms of the numerator (light-seconds, light-years, whatever), and the units of the $t$ axis in terms of the denominator (seconds, years, whatever). Velocities are then just some number $v$ of light-seconds per second, light-years per year, whatever, where the absolute value of $v$ is less than 1. That's really all there is to it.

 

[Mister T]

I think in this case c would have a number with units but does it really matter to have a number 1 without units if you achieve the same thing with units?

Note that c = 1 light-year per year. This is different from saying c=1 because in the former case you'd still need to include c in the equations to make the units come out right. If c=1 then you can omit c from the equations and they take on a much simpler form. So, instead of the famous $E_o=mc^2$ you simply have $E_o=m$. Or instead of $(mc^2)^2=E^2-(pc)^2$ you simply have $m^2=E^2-p^2$.

 

[vanhees71]

No, it's not different, because $1\text{ly}/{\text{ly}}=1$ ;-).

 

[Mister T]

No, it's not different, because $1\text{ly}/{\text{ly}}=1$ ;-).

My point is that $1\ \text{ly}/{\text{y}}$ is not the same as $1\ \text{ly}/{\text{ly}}$.

 

[JohnnyGui]

This isn't exactly wrong as you state it, but it appears to be leading you to confuse yourself. You are making this much more difficult than it needs to be. Drawing a spacetime diagram in units where $c = 1$ is just a convenient choice of units; that's all it is. If it helps, think of it as picking units so $c$ is 1 light-second per second, or 1 light-year per year, or something like that. Then you can just think about the units of the $x$ axis in terms of the numerator (light-seconds, light-years, whatever), and the units of the $t$ axis in terms of the denominator (seconds, years, whatever). Velocities are then just some number $v$ of light-seconds per second, light-years per year, whatever, where the absolute value of $v$ is less than 1. That's really all there is to it.

I think I'm starting to get it now. All they're trying to do in the diagram is to get the slope of light movement equal to 1 which means that for light x / t = 1. If we know that for light x = 3e^8 m for every 1 t, that means that the slope for light is normally x / t = 3e^8. So to make that equation result in a 1, one would have to mutiply the denominator t with c. Or, one could also divide x by c which gives (x / c) / t which can be rewritten again as x / (ct).

Am I making any sense?

 

[PeterDonis]

If we know that for light x = 3e^8 m for every 1 t,

For every 1 second of t. But that's because you chose the units of t to be seconds instead of meters--i.e., you chose the units of t so t is 1 for x = 3e^8 in the case of light. But you could have chosen units of t so that t = 3e^8 for x = 3e^8, so the speed of light would be 1. There is nothing in the physics that picks out units of seconds for t as being special; it's just a human convention.

So to make that equation result in a 1, one would have to mutiply t with c.

You're not "multiplying" t with c; you're changing the units of t so that t = 3e^8 for x = 3e^8 in the case of light. In other words, you're re-scaling the t coordinate of your coordinate system; you're not changing anything physical. It's no different than if you chose to use minutes instead of seconds for your time unit, or hours, or days, or whatever.

one could also divide x by c

You're not "dividing" x by c. You're changing the units of x so that x = 1 for t = 1 in the case of light; i.e., you're using units of light-seconds instead of meters for x. Once again, you're just re-scaling the x coordinate of your coordinate system; you're not changing anything physical. It's no different than if you chose to use feet instead of meters for your distance unit, or furlongs, or miles, or whatever.

 

[JohnnyGui]

You're not "multiplying" t with c; you're changing the units of t so that t = 3e^8 for x = 3e^8 in the case of light. In other words, you're re-scaling the t coordinate of your coordinate system; you're not changing anything physical. It's no different than if you chose to use minutes instead of seconds for your time unit, or hours, or days, or whatever.

But to change the units of t to make t= 3e^8 for x=3e^8, mustn't its units be changed into metres? And to change the units of t, which is s, into metres, mustn't s be multiplied by m/s which is velocity and thus be multiplied by c? I think that's what I was trying to say.

I have another question. While reading a wiki page on the axes of the Minkowski diagram I saw this example:

My question is, shouldn't the red arrow have an equation of x = 4ct (since it takes four times more ct units than light make the same x distance) and the green arrow x = 1/3 ct?
EDIT: Never mind about this, I was lost for a moment there.

 

[Ibix]

My question is, shouldn't the red arrow have an equation of x = 4ct (since it takes four times more ct units than light make the same x distance) and the green arrow x = 1/3 ct?

No. Write the equation of the red line as $x=mct$. Along that line, when $x=1$, $ct=4$. Therefore $1=m\times 4$, or $x=(1/4)ct$.

 

[JohnnyGui]

No. Write the equation of the red line as $x=mct$. Along that line, when $x=1$, $ct=4$. Therefore $1=m\times 4$, or $x=(1/4)ct$.

Wow, never mind about me asking this. I totally had some kind of a blackout here.

 

[Ibix]

It happens. The thing that probably tripped you is that the equations are written "back to front" here. You normally write the equation of a line as y=mx, but here they've written x=my because, with the vertical axis being ct, "m" is the velocity expressed as a fraction of the speed of light.

 

[JohnnyGui]

It happens. The thing that probably tripped you is that the equations are written "back to front" here. You normally write the equation of a line as y=mx, but here they've written x=my because, with the vertical axis being ct, "m" is the velocity expressed as a fraction of the speed of light.

That's exactly what tripped me. Probably a hint to take a 1 minute break.

 

[PeterDonis]

to change the units of t to make t= 3e^8 for x=3e^8, mustn't its units be changed into metres?

"Meters" of time, yes.

to change the units of t, which is s, into metres, mustn't s be multiplied by m/s which is velocity and thus be multiplied by c?

In the sense that you're rescaling coordinates, yes. Again, it's no different than if you had an event labeled with a $t$ coordinate of 1 second, and you decided to change your units of time to minutes, so you relabel that event to have a $t$ coordinate of 1/60 of a minute. Here you are changing your units of time to meters, so an event that was labeled with a $t$ coordinate of 1 second now gets relabeled to have a $t$ coordinate of $3 \times 10^8$ meters. But it's still the same event; nothing physical has changed. All that's changed is how you label events with coordinates.