# What does remain constant in relativity?

1. Nov 21, 2006

### y_stassin

Hello, I have just registered in the forum!
I am a shool student, who is generally inerested in Physics and especially in Relativity.

So I'm reading a fascinating book about Einstein's theory at this moment (it's in German and it's called Newton, Einstein und die Relativitaetstheorie of Harald Fritzsch).

And I've just learned, that although time and space depend on the relative movement of the observer, one thing (together with c) remains constant in every system. And this is the difference of the square of time- square of space. This is proved in the book with the help of the definition of the gamma factor. One should note that here the v used during this proof is v/c.
Then the book says the exact term of this difference should be (ct)^2 - x^2 .

normaly there is a term for the 3 dimensions. this is the one for the x-Direction

well... my question is: how can one prove that this is also constant? or in other words, why is this term the general one?

I hope you understood my problem

y_stassin

2. Nov 21, 2006

### Staff: Mentor

Actually, anything that can be expressed as the scalar product of two four-vectors is invariant between inertial reference frames. Your example is the scalar product of the position four-vector $(ct, x, y, z)$ with itself: $(ct)^2 - x^2 - y^2 - z^2$.

Another four-vector combines energy and momentum: $(E, p_x c, p_y c, p_z c)$. It's "square" is the square of the invariant mass (or "rest mass") of a particle or system, apart from factors of c: $E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2 = (m_0 c^2)^2$.

Further, the scalar product of these two four-vectors is invariant, that is, the quantity $Ect - p_x c x - p_y c y - p_z c z$, which doesn't have any special name as far as I know.

Last edited: Nov 21, 2006
3. Nov 21, 2006

### y_stassin

I'm having a bit of trouble inserting the formulas in the answer window... How did you do it? I use the Formel Editor in MS Word, but it cannot copy it in here.

Anyway I didn't understand quite well...:grumpy:

Isn't the skalar product of (ct,x,y,z) with itsself the thing you wrote, but with + instead of - ?? (I learned the skalar product just this year and I don't know if I'm right...)

Thanks again,

y_stassin

4. Nov 21, 2006

### robphy

In addition to the scalar quantities that jtbell mentioned, there are other invariant "geometrical" objects: tensors, where the scalar is the simplest type, and the vector is next simplest type. [Be careful not to confuse the components of a tensor [which aren't constant in every system] with the tensor itself.]

One way to answer the original question "how can one prove that this is also constant? or in other words, why is this term the general one?" is to write down the set of "transformations" (correspondences) that relate one system with another system. For special relativity, these are the Lorentz Transformations... which you may have seen in the simple case where the relative motion is in the x-direction.... but the general case involves relative motion in an arbitrary direction, possibly with a rotation.

There are also vector-algebraic ways to prove it without explicitly writing down the Lorentz Transformations (as a 4-by-4 matrix).... but I'm not sure if they would help you now.

5. Nov 21, 2006

### Staff: Mentor

Click on any equation and you'll get a window that shows how I entered it. You'll also see a link to a guide on how to use LaTeX.

For ordinary vectors the scalar product does have all + signs. For relativistic four-vectors (one timelike component and three spacelike components), we define the scalar product the way I used it, precisely because the resulting products are invariant between reference frames.

To show that (for example) $s^2 = (ct)^2 - x^2$ (in one spatial dimension) is invariant between reference frames, use the Lorentz transformation, which relates x and t in one reference frame with x' and t' in another:

$$x^\prime = \gamma (x - vt)$$

$$t^\prime = \gamma (t - vx/c^2)$$

where

$$\gamma = \frac{1}{\sqrt {1 - v^2 / c^2}}$$

Substitute these into $(ct^\prime)^2 - (x^\prime)^2$ and show that this reduces to $(ct)^2 - x^2$.

6. Nov 24, 2006

### daniel_i_l

The fact that an interval in ST between two events is constant can be proven using the fact that c is constant in every frame. if a photon leaves point A and goes to point B then the interval between those two events is 0! if it's not 0 then the speed isn't c, so since all observers see the same speed (c) they must see the same interval (0) two.

7. Nov 24, 2006

### lightarrow

Sorry, robphy, I'm not competent at all about tensors; which is the meaning of invariant tensor, if its components are not?

8. Nov 24, 2006

Staff Emeritus
Within the component formalism this statement is expressed that the tensors (here defined by components) are not invariant, but equations between tensors are. The big application of this is that if a tensor equals zero in one frame then it equals zero in all frames.

I think non-component tensors are equivalence classes of component tensors over diffeomorphisms; i.e, each one is the "orbit" of a given component-tensor in the group of diffeomorphisms. So they are invariant by definition.

9. Nov 24, 2006

### robphy

Here's an example from the weather [map].

At a given point on the map, we read a temperature, a scalar with one component T. If my friend has a rotated map (who has drawn a different set of axes on the map), my friend reads a temperature, a scalar with one component T'. Since temperature is a scalar, T'=T. We agree on the temperature.

At a given point on the map, we read a wind velocity vector, a vector with three components, vx, vy, vz. If my friend has a rotated map (who has drawn a different set of axes on the map), my friend reads a wind velocity vector, a vector with three components vx',vy',vz'. Since velocity is a vector, we will generally disagree with our components, vx'=/=vx, vy'=/=vy, vz'=/=vz . However, we will agree on its magnitude [its wind speed] and its direction [toward what city the wind blows]. Because of this, it better to represent the vector as an arrow (based at the given point) rather than "three axis-dependent numbers" (three axis-dependent scalars). The arrow is a more general view of this particular "geometric object": the vector.

10. Nov 25, 2006

### lightarrow

Thank you robphy; thank you selfAdjoint.

11. Nov 25, 2006

### bernhard.rothenstein

pedagogy of helping

My experience posting on the Forum enables me to share some thoughts with my distinguished coleagues on the basis of that thread and on the answers on it.
The solicitor is a student reading an introductory textbook. The help he is asking for is not very clear. Ifs he interested how to go from one space dimension to three ones? In that case I would advise him to go to the well known thought experiment which involves two rods perpendicular to the direction of relative motion in relative motion that does not involve light signals involving rather a piece of chalk.
If he is interested in the derivation of xx-cctt I would advise him to start with the synchronization of two clocks (x,0) and (-x,0) using a light signal emitted from the origin O at t=0. After synchronixation the readings of the two clocks are t and it is obvious that x=(+/-)t where from we obtain that
xx-cctt=0 The invariance of y and z demonstrated by the experiment mentioned above enables us to extend it to xx+yy+zz-cctt=0. Considering the same experiment from another inertial reference frame moving relative to the first one we obtain that x'x'+y'y'+z'z'-cct't'=0 and the way is paved for the poster of the thread.
Sending him to four vectors or to tensors is a premature advise.
A teacher of mine told me that mathematics in physics is like raisin in the cake. But cake is good without raisins. Of course to mutch raisins make the cake hard to digest. Some times less can be more!
I have noticed such situations in the case of many threads.
The point of view of the student who posted the thread would be of help.
sine ira et studio