# Lorentz Transformation or Time Dilation? [closed]

Say there is a spaceship really far away from Earth moving at 80% the speed of light (away from Earth-edited), at some point a radio signal is sent by observers on Earth. I need to be able to calculate how long it would take for the signal to reach the spaceship. Would calculating the time required for the signal to reach the spaceship in Earth's frame, which is pretty straight forward and then applying time dilation directly be valid? If not please explain what the issue behind the approach is. The other approach would be to apply Lorentz transformations, but honestly I don't get what exactly makes one approach right and other one wrong in certain scenarios.

• Your question cannot be answered because you have omitted key words that an answer would need. There is a good reason why the name of the theory is Relativity. Oct 3 at 15:06
• You'd need more information. If the ship is moving towards the earth, away, at an angle, makes a difference. Time dilation is part of the Lorentz transform, the Lorentz transform is more general Oct 3 at 16:09

Simple time dilation is not sufficient since the space between the spaceship and the Earth is distorted in the frame of the spaceship relative to the Earth frame. As a general rule, a Lorentz transformation will always work; simple length or time dilation are special cases of the full linear transformation.

Suppose an event (i.e. radio signal reaches spaceship in Earth's frame) occurs at coordinates $$(t, x, y=0,z=0)$$). For a boost to frame with $$\beta = v/c$$ along $$x$$:

$$\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} ct \\ x \\ 0 \\ 0 \end{bmatrix}$$ $$= \begin{bmatrix} \gamma(ct - \beta x) \\ \gamma(x - \beta ct) \\ 0 \\ 0 \end{bmatrix}$$

So in the frame of the spaceship the signal arrives at $$t' = \frac{\gamma}{c}(ct - \beta x)$$, $$x' = \gamma(x - \beta c t)$$. Here you can clearly see that simple time dilation would not be correct.

Another helpful way to understand these transformations is to use Minkowski diagrams, which are basically the graphical equivalent of this matrix transformation. There are plenty of other resources online to help you try to understand how these work.

• "a Lorentz transformation will always work" +1 for this. IMO, time dilation and length contraction formulas should not be taught to new students. They automatically drop out of the Lorentz transform when appropriate
– Dale
Oct 3 at 15:10
• @Dale Agreed. I think the issue is that the individual formulas feel more intuitive, so they're easier to teach. The issue is that, as soon as you start getting slightly more complex situations, the intuition in special relativity becomes extremely difficult. Oct 4 at 16:44

The time dilation equation only applies in a very limited set of circumstances, namely when you are comparing the elapsed time between two events that occur in the same place in one frame with the corresponding elapsed time in another frame in which they occur in different places. For example, if you wanted to calculate the time that passes on the spaceship between leaving earth and arriving at a distant planet, you could use the time dilation equation. However, the problem you have described doesn't fit that pattern, so the time dilation equations doesn't apply.

• I see, in the scenario you mentioned to get an intuitive understanding it would be right to consider two ticks of a clock in spaceship as separate events and as clock is at rest wrt spaceship it would be right to apply time dilation. In the scenario I mentioned the radio signal is initially at Earth and then reaches the spaceship, the signal leaving Earth could be considered as one event and then reaching the spaceship could be considered as another event. Since the displacement between the events is not zero in neither frames it would be wrong of me to apply time dilation. Am I right? Oct 3 at 17:49
• That is exactly right. To use the simple time dilation formula, the interval must span two events in the same location in one of the two frames being compared. Oct 3 at 19:09

A problem with applying time dilation willy nilly is that you need to know where ship is when the signal is sent, in ship's frame.

If you call the Earth (Ship) frame $$S$$ ($$S'$$) and write events as $$(t, x)$$ and $$(t', x')'$$, with the signal sent at:

$$E_{Tx} = (0, 0) = (0, 0)'$$

(note the ship is not at the origin of its own frame).

In $$S$$, the world line of the ship is ($$c=1$$):

$$P(t) = (t, D + \beta t)$$

and the transmitted signal's is:

$$T(t) = (t, t)$$

The receive event occurs when they overlap:

$$T(t) = P(t)$$

for which we can solve the spatial component:

$$t_{Rx} = D + \beta t_{Rx}$$ $$t_{Rx} = \frac D {1-\beta}$$

so:

$$E_{Rx} = (\frac D {1-\beta},\frac D {1-\beta})$$

A Lorentz Transform to $$S'$$ gives:

$$t'_{Rx} =\gamma(\frac D {1-\beta} - \beta \frac D {1-\beta})= \gamma D$$

That may seem unexpected.

First, note the ratio of the elapsed times is:

$$R = \frac{1/(1-\beta)}{\gamma} = \frac{1/(1-\beta)} {1/\sqrt{1-\beta^2}}$$

$$R = \frac{1-\beta^2}{1-\beta} = \frac{\sqrt{(1-\beta)(1+\beta)}}{1-\beta}$$

$$R = \sqrt{\frac{1+\beta}{1-\beta}} = f_D$$

which turns out to be relativistic Doppler shift factor.

Finally: to resolve this confusion, you need to really understand:

"That no inherent meaning can be assigned to the simultaneity of distant events is the single most important lesson to be learned from relativity."

— David Mermin, It’s About Time

The question is, where is the space ship in the Earth frame at $$t'=0$$, when the signal is transmitted?

It's not at $$(0, D)$$...which is where it is in the Earth frame.

I'll leave the answer to that question as an exercise.