# Lorentz vs space time interval

• ersteller
In summary, the conversation discusses the relationship between the Lorentz transform and the invariance of the space time interval. It is shown that using the Lorentz transform, one can find the time interval t' measured by system S', which is moving with velocity v with respect to system S. However, there is also an inconsistency in using the invariance of the space time interval, as the calculation results in a missing gamma factor. This is due to a common misconception about length contraction, where events that are simultaneous in one frame are not necessarily simultaneous in the other frame. Overall, the conversation highlights the need for understanding the concept of simultaneity in order to correctly apply the Lorentz transform and invariance of the
ersteller
Hallo.
A question about the relationship between the formulas found using the Lorentz transform and the invariance of the space time interval.

Two events A and B occur at the same time and different space locations in system S, where A and B are at rest and at distance x.
The system S' moves with velocity v with respect to S.

I want to find the time interval t' that S' measures between A and B.

1. By using the Lorentz transform one finds immediately:

$$t' = -\gamma\frac{v}{c^{2}}x$$

2. By using the invariance of the space time interval:

$${(ct')^{2} - x'^{2} = - x^{2}}$$

and using the contacted $$x' = \frac{x}{\gamma}$$

one finds (neglecting the negative square...)

$$t' = -\frac{v}{c^{2}}x$$

So here the gamma is missing... Must I understand that:

a.
With the Lorentz transform I find the value of the time interval 'assigned' at S' as measured by S (with the time dilation due to the relative velocity between S and S')

and

b.
With the invariance of the space time interval I find this time interval as measured by S' (analogously to the use of the contracted x' in the expression of the interval)?

Have I right understood?
Thank you
er

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I don't quite understand the wording of your question at the end, but think I can see where the anomaly in your calculation comes from. (NOTE: I'll use units where c = 1 to make the equations simpler; c can be replaced afterwards, using dimensional analysis.)

If we use the Lorentz transformation, setting t = 0,

$$t'=\gamma(t-vx)=-\gamma v x$$

just as you found. So far so good... And, again setting t = 0,

$$x'=\gamma(x-vt)=\gamma x$$

$$x^2=(x')^2-(t')^2$$

$$(t')^2=(x')^2-(x')^2\gamma^{-2}=(x')^2(1-(1-v^2))$$

$$(t')^2=v^2(x')^2=\gamma^2 v^2 x^2$$

$$(t')=\pm \gamma x v$$

You got inconsistent results because you put $x$ and $(x')^2$ the wrong way round in the equation after you wrote "and using the contacted".

Thank you; have you please the patience to help me again?

My aim is to find t' using the invariance of the space time interval.
This interval calculated in S is x2.
This interval in S' should be x'2 - t'2.
It should be x'2 - t'2 = x2.

S' moves wrt S at velocity v, so, as I qualitatively understand, it should be x'<x.

You obtained the right result by putting $$x' = \gamma x > x$$.

I think I have here a HUGE misunderstanding !

Thanks!
er

This is a common misconception. I had the same problem when I first learned it. You are thinking that x' should be less than x, because of length contraction. This is incorrect. In fact, for the case you stated x' >x, as Rasalhague showed. But then what about length contraction? How does that work? To understand this, you need to understand that events that are simultaneous in one frame are not necessarily simultaneous in the other frame. So, consider a rod of length L in the unprimed frame. The two ends of the bar have the following unprimed coordinates ( all coordinates are (t,x)):

$$E1 = (0,0) E2 = (0,L)$$

So in the unprimed frame, the length of the bar is the difference of the two x-coordinates, which is L. Now, transform these into primed coordinates using the Lorentz transformation, and you find:

$$E1' = (0,0) E2' = ({\beta} {\gamma} L, {\gamma} L)$$

Now, the length of the bar in the primed frame is not just the difference in the x coordinates, because the two events are now at different times . To find the length of the bar in the primed frame, I need to find coordinates of the two ends at the same value of t'. You can do this by asking what time in the unprimed frame gives a value of t' = 0 in the primed frame. These new events will be:

$$E1= (0,0) E3 = (-\beta L,L) E1' = (0,0) E3' = (0,\gamma (1-\beta^2)L) = (0,\frac{L}{\gamma})$$

Now we can find the length of the bar in the primed frame by just taking the difference of the x-coordinates between E1' and E3', which are now at the same time in the primed frame. As expected, now L'<L, indicating length contraction. Does this help?

Thank you phyzguy, it helps a lot!

Now I've understood my mistake and I can go on, and find other mistakes...

For example, from

$${(ct')^{2} - x'^{2} = - x^{2}}$$

can one really find t' without using the Lorentz transformations? I thought yes, but how to?

Ciao e grazie
er

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## 1. What is the difference between Lorentz and space-time interval?

Lorentz and space-time interval are two different concepts used in the theory of relativity. Lorentz refers to the transformation equations that describe how measurements of space and time change between two reference frames in relative motion. Space-time interval, on the other hand, is a measure of the distance between two events in space and time, and is used to calculate the space-time interval between two points in a reference frame.

## 2. How is Lorentz related to special relativity?

Lorentz transformations were developed by Hendrik Lorentz and used by Albert Einstein in his theory of special relativity. These transformations explain how measurements of space and time change between two observers in relative motion. They are a key component of special relativity and help to reconcile the laws of physics with the principle of relativity.

## 3. Can you use Lorentz equations to describe motion in both space and time?

Yes, Lorentz transformations can be used to describe both spatial and temporal measurements. They take into account the fact that measurements of both space and time are relative and can change depending on the observer's frame of reference. This is a fundamental concept in special relativity.

## 4. How is the space-time interval calculated?

The space-time interval is calculated using the Pythagorean theorem, where the distance in space and time are represented by the sides of a right triangle. The equation for calculating the space-time interval is Δs² = Δx² + Δy² + Δz² - (cΔt)², where c is the speed of light. This equation takes into account the fact that space and time are intertwined and cannot be considered separately.

## 5. What is the significance of the space-time interval in physics?

The space-time interval is significant in physics because it is a fundamental concept in special relativity. It allows us to understand how measurements of space and time change between different reference frames and how this affects the laws of physics. It also helps to explain phenomena such as time dilation and length contraction, which have been confirmed through various experiments and observations.

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