# How does the time dilation equation work?

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1. Jan 3, 2016

### Isaac0427

Hi all!
I've been doing some studying length contraction and time dilation and I came across this link http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html . I completely understand what it did with length contraction but when I got to time dilation I couldn't figure out one thing. They appeared to be using the equation t=(t'+vx'/c2)γ, but I thought that the equation was t'=(t+vx/c2)γ. Using that equation I got T=T0/γ and not T=T0γ, but I know that is wrong. Can you please help me understand this?
Thanks!

2. Jan 3, 2016

### bcrowell

Staff Emeritus
To invert the Lorentz transformation, you just flip the sign of v. So the two equations you wrote are equivalent except that to make them consistent with each other you would need to replace v with -v in one of them.

3. Jan 3, 2016

### Isaac0427

Ok, but then wouldn't you have to flip the sign of the velocity in the γ term?

4. Jan 3, 2016

### Mister T

You, and the article you linked, are not using consistent notation. The proper time is the time elapsed in a frame of reference where the change in position is zero. Let's start with your equation $t'=(t+\frac{vx}{c^2}) \gamma$. If we set $x=0$ that makes $t$ a proper time (assuming a reference event at (0,0), as usual). Thus we have $t'=t \gamma$.

Note that the transformation equation you wrote assumes the primed frame is moving to the left relative to the unprimed frame, which is not the usual convention. The usual convention is the opposite, so that $t'=(t-\frac{vx}{c^2}) \gamma$, but in this case it doesn't matter.

And changing the sign of $v$ does not change the sign of $\gamma$.

5. Jan 3, 2016

### Staff: Mentor

They're same equation except that they're saying t' and x' where most people would say t and x, and vice versa. You could do something similar with the quadratic formula from first-year algebra that says that the solution to the equation $ax^2+bx+c=0$ is $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$; just say that the solution to the equation $cx^2+bx+a=0$ is $\frac{-b\pm\sqrt{b^2-4ac}}{2c}$ - it looks different but that's just because we're playing games with the names of two of the variables.

If you flip which coordinates are the primed ones, you also have to flip which time interval is labeled $T$ and which is $T_0$. You ended up with the right result using the weird convention in which $T$ and $T_0$ have been flipped.

6. Jan 4, 2016

### Isaac0427

Ok, so first of all, there was a mistake in my original post. I said t'=t+vx/c2 and x'=x+vt instead of t'=t-vx/c2 and x'=x-vt. Using this here's my thinking:
Let T=t2-t1, T0=t'2-t'1, L=x2-x1 and L0=x'2-x'1. Using this and the Lorentz transformations, I got T0=Tγ assuming x1=x2 and L0=Lγ assuming t1=t2, which would also give me T=T0/γ and L=L0/γ.
What did I do wrong?

7. Jan 4, 2016

### Isaac0427

Ok, so after reviewing this it appears to me that you get a different answer if you use the regular Lorentz transformation than if you use the inverse Lorentz transformation. Is this correct? If yes, how do you know which one to use?

8. Jan 4, 2016

### Mister T

Proper time is the time elapsed in the frame where the change in position is zero.

9. Jan 4, 2016

### PeroK

It's great you're learning SR, but I'd say you should try to sharpen up your mathematical thinking a little. You know which transformation to use from the way you set up your problem. You must choose a direction to be positive. (Keeping this to motion in one dimension). Then, if you have an object moving, it has either a positive or negative velocity. Likewise, a reference frame can be moving in the positive or negative directions.

Any formula, including the Lorentz Transformation (LT), depends on these things. You can't just apply any formula without checking that the formula applies to the way you have set up your problem. It's no different from, say, knowing when to take gravity to be positive (when down is the positive direction) and negative (when up is the positive direction).

Take a look at the way the LT is derived/defined and note the assumptions that relate one frame (normally unprimed) with another frame (normally primed, S' or whatever). If you do that, you'll see when to use the LT and when to use the "inverse" LT.

10. Jan 4, 2016

### Isaac0427

So if I am understanding you correctly, it is completely based on the situation. I was just wondering if there is some mathematical reason to use one or the other.

11. Jan 5, 2016

### PeroK

The usual Lorentz Transformation is:

$t' = \gamma (t - vx/c^2)$ and $x' = \gamma (x - vt)$

This relates the coordinates of time $t'$ and distance $x'$ for a frame moving at velocity $v$ with respect to another frame. There are three important points: the frames have a common origin: $t = 0, x = 0$ corresponds to $t' = 0, x' = 0$; the positive $x/x'$ direction is the same for both frames; and, $v$ is positive if the motion of the "moving" frame is in the positive x-direction.

For example. If $v = c/2$ then this means the moving frame is moving to the right (positive x-direction), In this case you have:

$t' = \gamma (t - x/2c)$ and $x' = \gamma (x - ct/2)$

The "inverse" LT is:

$t = \gamma (t' + vx'/c^2)$ and $x = \gamma (x' + vt')$

This gives the $t, x$ coordinates in terms of $t', x'$. In the same set-up as above.

Now, you can see this two ways. The quick way is to note that you can consider the unprimed frame as moving to the left with respect to the primed (') frame. So, you just apply the normal LT equation using $-(-v) = +v$ instead of $-v$ and swap the roles of the coordinates (my first suggestion is that you think that through).

For example. If $v = c/2$ you have:

$t = \gamma (t' + x'/2c)$ and $x = \gamma (x' + ct'/2)$

The second way is to do the algebra and rearrange the LT equations to express $t', x'$ in terms of $t, x$. This will confirm the quick way. My second suggestion is that you do that too. It's a good algebraic exercise if nothing else.

You can see now in the hyperphysics page that they were simply using the inverse LT to relate coordinates in the "moving" frame to coordinates in the "rest" frame. And that the set-up on that page is as I described it above.

12. Jan 5, 2016

### Isaac0427

Ok, I completely understand this. My only question is why do you use the regular transformation for length contraction and the inverse transformation for time dilation.

13. Jan 5, 2016

### PeroK

It depends what they are trying to do. That page doesn't have much explanation, so it's maybe not a good place to learn. Especially the length contraction needs a bit more explanation. Over to Mister T ...

14. Jan 5, 2016

### Mister T

You can use either transformation to derive either effect. You originally asked about a confusion over time dilation, so let's look at that. (Assume a reference event at $(0,0)$ as usual).

Consider the transformation equation $t'=(t-\frac{vx}{c^2}) \gamma$. If I let $x=0$ then $t$ is by definition a proper time and $t'=t \gamma$.

Consider the inverse transformation equation $t=(t'+\frac{vx'}{c^2}) \gamma$. If I let $x'=0$ then $t'$ is by definition a proper time and $t=t' \gamma$.

In all cases you multiply the proper time by $\gamma$ to get the dilated time. Since $\gamma \geq 1$ the dilated time is greater than or equal to the proper time.

15. Jan 5, 2016

### Isaac0427

Oh, so depending on which transformation you use, either L or L0 could be proper time. This makes more sense. Thank you.

16. Jan 5, 2016

### Mister T

You can use whatever you want to refer to whatever you choose, but keep in mind that when you use nonstandard notation you have trouble communicating with others. Usually $L$ is used for a distance or length, not a time. And usually the subscript "o" refers to the proper value. So $L_o$ would be a proper length and $t_o$ would be a proper time. It's even more common to call $\tau$ the proper time.

The link in your first post is sloppy and inconsistent with its notation, first using $\tau_o$ for the proper time and then later switching to $T_o$. This is at least a factor in your confusion if not the outright cause.

In your first post you first used $t$ and $t'$ and then later switched to $T_o$ and $T$ without any indication, let alone a clear indication, of what that change in notation meant. All of your confusion is based on the absence of that being sorting out.

Last edited: Jan 5, 2016
17. Jan 6, 2016

### PeroK

I don't like the explanation for length contraction on that page. Here's how I would look at it.

Imagine we have an object moving at velocity $v$. At $t=0$ the ends are at $x_1$ and $x_2$. At time $t$, therefore, the ends are at:

$(t, x_1 + vt)$ and $(t, x_2 + vt)$

And the length of the object in this frame is $x_2 - x_1$

What are the coordinates of the two ends in a frame moving with the object at velocity $v$?

$x_1' = \gamma (x_1 + vt - vt) = \gamma x_1$

$x_2' = \gamma (x_2 + vt - vt) = \gamma x_2$

The object is at rest in this frame (its position does not depend on $t'$) and so the length of the object is measured as:

$x_2' - x_1' = \gamma (x_2 - x_1)$

18. Jan 6, 2016

### Isaac0427

In this example x2-x1 is proper length. If the equations were
$x_1 = \gamma (x_1' + vt' - vt') = \gamma x_1$
$x_2 = \gamma (x_2' + vt' - vt') = \gamma x_2$
$x_2 - x_1 = \gamma (x_2' - x_1')$
then x'2-x'1 would be proper length, correct?

19. Jan 6, 2016

### Isaac0427

Sorry, I meant T and T0, not L and L0.

20. Jan 6, 2016

### PeroK

Not at all. Proper length is the length in a frame where the object is at rest.