I Deriving Lorentz from time dilation and length contraction

[zox00]

Is it possible to derive the Lorentz transformation from time dilation and length contraction?
If so, how should I start?
I know how to derive it while considering 4 scenarios finding values of A, B,D,E in x'=Ax+Bt t'=Dx+Et
and the transformation is:
x'=(x-vt)/sqrt(1-v^2/c^2)
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
y'=y
z'=z

Thanks beforehand :)

 

[Ibix]

As you note, there are four constants to be determined in the general linear transform. Assuming the time dilation and length contraction formulae fixes two of them. You need some additional facts.

 

[jtbell]

Hint: which phenomenon often needs to be invoked in order to resolve relativistic "paradoxes" such as the "barn and pole"?

 

[PeroK]

Is it possible to derive the Lorentz transformation from time dilation and length contraction?

No, it's not. Something else is missing.

 

[zox00]

I asked the teaching assistant to clarify it. We can derive time dilation and length contraction from Lorentz transformation. But for this question: He said we need to do the process backward and find Lorentz transformation formulas for x' and t' (x'=(x-vt)γ, t'=(t-vx/c^2)γ) from time dilation and length contraction formulas which are x'=x/γ, t'=t*γ

 

[PeterDonis]

He said we need to do the process backward and find Lorentz transformation formulas for x' and t' (x=(x-vt)γ, t'=(t-vx/c^2)γ) from time dilation and length contraction formulas which are x'=x/γ, t'=t*γ

This is not possible. The Lorentz transformation also includes relativity of simultaneity; without that you can't do what you are being asked to do.

 

[zox00]

This is not possible. The Lorentz transformation also includes relativity of simultaneity; without that you can't do what you are being asked to do.

Here's the full question:
Derive the Lorentz transformation starting from time dilation and length contraction formulas. Do you need to make any assumptions? Why? Why not? Briefly discuss the implications of your derivation.

 

[PeterDonis]

Here's the full question

Well, the full question asks if you need to make any assumptions. So clearly it allows for the possibility that you will need to make some, i.e., that time dilation and length contraction alone might not be sufficient to get the full Lorentz transformation equations.

 

[zox00]

Well, the full question asks if you need to make any assumptions. So clearly it allows for the possibility that you will need to make some, i.e., that time dilation and length contraction alone might not be sufficient to get the full Lorentz transformation equations.

Yes, it does. Any help on what those assumptions could be?

 

[stevendaryl]

Sticking to just two spacetime coordinates, $x$ and $t$, the following linear transformation satisfies time dilation and length contraction:

• $x' = \gamma (x - v t)$
• $t' = \frac{1}{\gamma} t - A (x - v t)$

The constant $A$ is arbitrary.

To see that it satisfies time dilation, let the events $e_1$ and $e_2$ take place at the same value of $x'$, with $e_2$ occurring after $e_1$. Let $(x_1, t_1)$ be the coordinates of $e_1$ in the unprimed frame and let $(x_1', t_1')$ be the coordinates in the primed frame. Similarly, let $(x_2, t_2)$ and $(x_2', t_2')$ be the coordinates of $e_2$ in the two frames. Let $\delta x = x_2 - x_1$, etc. Then the Lorentz transformations tell us

$\delta x' = \gamma (\delta x - v \delta t)$
$\delta t' = \frac{1}{\gamma} \delta t - A (\delta x - v \delta t)$

Since $\delta x' = 0$, that implies that $\delta x - v \delta t = 0$. So we have: $\delta t' = \frac{1}{\gamma} \delta t$. So the clocks at rest in the primed frame run slower.

To see that it satisfies length contraction, consider a rod at rest in the primed frame of length $L$ as measured in the primed frame. Let $e_1$ be the measurement of one end of the rod in the unprimed frame, and let $e_2$ be the measurement of the other end. For it to be a valid length measurement, $e_1$ and $e_2$ must take place at the same time, according to the unprimed frame. That means that $\delta t =0$.

$\delta x' = \gamma (\delta x - v \delta t)$. Using $\delta x' = L$ and $\delta t = 0$ gives:

$L = \gamma \delta x$

So $\delta x = L/\gamma$. So the rod is shorter when measured in the unprimed frame.

So the additional term $A (x - vt)$ in the formula for $t'$ has no effect on time dilation or length contraction. So its value is arbitrary.

To pin down the value of $A$, you need another constraint on the transformation. Any of the following will do:

1. Assume that an object at rest in the unprimed frame is traveling at velocity $-v$ as measured in the primed frame.
2. Assume that light has the same speed $c$ in every frame.
3. Assume that the inverse transformation has the same form as the forward transformation (except for the substitution $v \rightarrow -v$.

 

[zox00]

Sticking to just two spacetime coordinates, $x$ and $t$, the following linear transformation satisfies time dilation and length contraction:

• $x' = \gamma (x - v t)$
• $t' = \frac{1}{\gamma} t - A (x - v t)$

The constant $A$ is arbitrary.

To see that it satisfies time dilation, let the events $e_1$ and $e_2$ take place at the same value of $x'$, with $e_2$ occurring after $e_1$. Let $(x_1, t_1)$ be the coordinates of $e_1$ in the unprimed frame and let $(x_1', t_1')$ be the coordinates in the primed frame. Similarly, let $(x_2, t_2)$ and $(x_2', t_2')$ be the coordinates of $e_2$ in the two frames. Let $\delta x = x_2 - x_1$, etc. Then the Lorentz transformations tell us

$\delta x' = \gamma (\delta x - v \delta t)$
$\delta t' = \frac{1}{\gamma} \delta t - A (\delta x - v \delta t)$

Since $\delta x' = 0$, that implies that $\delta x - v \delta t = 0$. So we have: $\delta t' = \frac{1}{\gamma} \delta t$. So the clocks at rest in the primed frame run slower.

To see that it satisfies length contraction, consider a rod at rest in the primed frame of length $L$ as measured in the primed frame. Let $e_1$ be the measurement of one end of the rod in the unprimed frame, and let $e_2$ be the measurement of the other end. For it to be a valid length measurement, $e_1$ and $e_2$ must take place at the same time, according to the unprimed frame. That means that $\delta t =0$.

$\delta x' = \gamma (\delta x - v \delta t)$. Using $\delta x' = L$ and $\delta t = 0$ gives:

$L = \gamma \delta x$

So $\delta x = L/\gamma$. So the rod is shorter when measured in the unprimed frame.

So the additional term $A (x - vt)$ in the formula for $t'$ has no effect on time dilation or length contraction. So its value is arbitrary.

To pin down the value of $A$, you need another constraint on the transformation. Any of the following will do:

1. Assume that an object at rest in the unprimed frame is traveling at velocity $-v$ as measured in the primed frame.
2. Assume that light has the same speed $c$ in every frame.
3. Assume that the inverse transformation has the same form as the forward transformation (except for the substitution $v \rightarrow -v$.

Many Thanks for the answer but I couldn't relate it to my problem :/
I'm not sure how should I get the best of what you've written.
I need to derive Lorentz from time dilation and length contraction.
Here's the full question:
Derive the Lorentz transformation starting from time dilation and length contraction formulas. Do you need to make any assumptions? Why? Why not? Briefly discuss the implications of your derivation.

 

[PeterDonis]

Many Thanks for the answer but I couldn't relate it to my problem :/

Try harder. @stevendaryl has already given you a good portion of the answer; notice that the first of the two equations at the top of his post is the Lorentz transformation equation for $x'$.

If his use of $\delta x$ and $\delta t$ confuses you, think of his event 1 as being at the spacetime origin $(0, 0)$ in both frames. The key thing is to understand why he included two events in his post instead of just one. Think about what "length contraction" and "time dilation" actually mean: are they properties of single events? Or are they properties of pairs of events (and if so, which pairs)?

 

[zox00]

Try harder. @stevendaryl has already given you a good portion of the answer; notice that the first of the two equations at the top of his post is the Lorentz transformation equation for $x'$.

If his use of $\delta x$ and $\delta t$ confuses you, think of his event 1 as being at the spacetime origin $(0, 0)$ in both frames. The key thing is to understand why he included two events in his post instead of just one. Think about what "length contraction" and "time dilation" actually mean: are they properties of single events? Or are they properties of pairs of events (and if so, which pairs)?

I appreciate it but I have no clue how I should proceed. I believe it must be something much simpler than this.

 

[stevendaryl]

Many Thanks for the answer but I couldn't relate it to my problem :/
I'm not sure how should I get the best of what you've written.
I need to derive Lorentz from time dilation and length contraction.
Here's the full question:
Derive the Lorentz transformation starting from time dilation and length contraction formulas. Do you need to make any assumptions? Why? Why not? Briefly discuss the implications of your derivation.

What is it that you don't understand?

 

[Herman Trivilino]

I believe it must be something much simpler than this.

Why?

 

[zox00]

Why?

Because the assistant said so :)
We derive length contraction from Lorentz, now we are supposed to derive lorentz from length contraction. I need to somehow traceback the formula of length contraction to Lorentz.

 

[zox00]

What is it that you don't understand?

It seems to me you derived length contraction from Lorentz.

 

[stevendaryl]

It seems to me you derived length contraction from Lorentz.

No. I gave an example of a transformation that was NOT the Lorentz transformation, but it had length contraction and it had time dilation. So that shows that the Lorentz transformations are not the only transformations that have length contraction and time dilation.

 

[stevendaryl]

The fact that it's not possible to derive the Lorentz transformations from length contraction and time dilation is a counting argument:

If you start with a general linear transformation:

$x' = A x + B t$
$t' = D x + E t$

there are 4 constants to figure out: $A, B, C, D$. To solve for 4 unknowns you need 4 independent equations.

Assuming that distances for the primed system are contracted by a factor of $\gamma$ gives you one equation. Assuming that time for the primed system is dilated by a factor of $\gamma$ gives you another equation. So you need 2 more equations somehow.

 

[PeterDonis]

Because the assistant said so :)

Did the assistant say you don't need to make any additional assumptions? Note that "additional assumptions" is the same thing as the "2 more equations" that @stevendaryl is referring to: making additional assumptions allows you to impose additional equations that constrain the transformation.

 

[zox00]

The fact that it's not possible to derive the Lorentz transformations from length contraction and time dilation is a counting argument:

If you start with a general linear transformation:

$x' = A x + B t$
$t' = D x + E t$

there are 4 constants to figure out: $A, B, C, D$. To solve for 4 unknowns you need 4 independent equations.

Assuming that distances for the primed system are contracted by a factor of $\gamma$ gives you one equation. Assuming that time for the primed system is dilated by a factor of $\gamma$ gives you another equation. So you need 2 more equations somehow.

Yes I don’t have to start from there. I don’t have to solve these 4 equations

 

[zox00]

Did the assistant say you don't need to make any additional assumptions? Note that "additional assumptions" is the same thing as the "2 more equations" that @stevendaryl is referring to: making additional assumptions allows you to impose additional equations that constrain the transformation.

From the question it seems like I need to make assumptions but I don’t have to do the transformation from
X’=Ax+Bt
T’=Dx+Et

I just need to trace back the derivation of length contraction from Lorentz transformation and go back to Lorentz

 

[Ibix]

From the question it seems like I need to make assumptions but I don’t have to do the transformation from
X’=Ax+Bt
T’=Dx+Et

I just need to trace back the derivation of length contraction from Lorentz transformation and go back to Lorentz

Yes you do. You may be disguising it somehow, but the underlying fact is that you need to identify a specific linear transform out of all possible linear transforms, and to do that you have to specify all four of those constants somehow. Assuming the time dilation and length contraction formulae only gives you two constraints. You need two more.

I just need to trace back the derivation of length contraction from Lorentz transformation and go back to Lorentz

Try it. You'll find that the forward process makes steps you can't reverse without making assumptions.

 

[PeterDonis]

I don’t have to solve these 4 equations

You came here asking for advice. We've given it to you. If you don't want to listen to it, that's up to you, but there's no point in a discussion if you're just going to ignore what we tell you.

Thread closed.