# Time like space like

1. Aug 8, 2007

### bernhard.rothenstein

Please tell me when is a physical quantity time like or space like

2. Aug 8, 2007

### robphy

A 4-vector is time- [alt: space-] like is when it points inside [alt: outside] the light-cone [in the tangent space]... as defined in Minkowski's famous paper.

3. Aug 8, 2007

### bernhard.rothenstein

time like -space like

Thanks. Have physical quantities which transform as a space coordinate transforms (say momentum) or physical quantities which transform as time transforms (say mass) have specific names?

4. Aug 8, 2007

### robphy

Some 4-vectors are seen to be of one type... like a 4-velocity is always timelike and the 4-acceleration is always spacelike [i.e. orthogonal to a timelike vector].

If you are trying to work with [coordinate-dependent] parts of a 4-vector,
you may be looking for a description like "the timelike- and spacelike-parts of a 4-vector.. with respect to choice of unit timelike vector" (akin to parallel- and perpendicular- components of a vector...with respect to a choice of unit vector).

5. Aug 8, 2007

### Meir Achuz

"which transform as time transforms (say mass)"
Ugh. Mass is a scalar invariant.
One can speak of the "time component" or the "space components" of a four-vector, and most physicists will know what is meant.

6. Aug 23, 2007

### bernhard.rothenstein

mass is a scalar invariant

I have posted on the Forum:

A short relativistic story but what is its moral?

--------------------------------------------------------------------------------

When I started to learn special relativity I knew that in a given inertial reference frame I the mass m, the speed u and the momentum p of a given particle are related by
p=mu (1)
Special relativity tought me that in an inertial reference frame I' which moves with speed V relative to I in the positive direction of the overlapped axes OX(O'X') (1) should read
p'=m'u'. (2)
Less or more complicated derivations lead to the following transformation equations for momentum and mass (g(V)=1/sqrt(1-V^2/c^2)
p=g(V)p'(1+V/u') (3)
m=g(V)m'(1+Vu'/c^2) (4)
"Old fashioned" physicists say that (4) relates the "relativistic mass" m and the "relativistic mass m') of the same particle measured by observers from I and I' respectively. If the particle is at rest in I' (u'=0) observers of that frame measure its "rest mass m(0) and (4) leads to
m=g(V)m(0) (5)
Taking into account that c has the same magnitude in all inertial reference frames in relative motion, all transformation equations remain "relativistically correct" if we multiply both their sides by a power of c. Doing so with (4) we obtain nothing interesting because mc and m'c have no physical meaning (no tardyon can move with speed c). Multiplying both sides 0f (4) with c^2 leads to E=mc^2 and E'=m'c^2 respectively which has the physical dimensions of energy (4) becoming
E=g(V)E'(1+Vu'/c^2)=g(V)(E'+Vp') (6)
(3) becoming
p=g(V)(p'+VE'/c^2). (7)
E=g(V)E(0) (8) E(0) representing the rest energy.
Presenting the transformation equations as (6), (7) and(8) the "new generation" of relativists have nothing to comment.
Is the dispute between the generations solved? Are the frenzied debates motivated?
soft words and hard arguments please

but nobody reacted. You will?

7. Aug 25, 2007

### pmb_phy

Hi Bernhard

>I have posted on the Forum
>When I started to learn special relativity I knew that in a given inertial reference frame I the mass
>m, the speed u and the momentum p of a given particle are related by
>
>p=mu (1)

Gotcha so far.

> Special relativity tought me that in an inertial reference frame I' which moves with speed V
>relative to I in the positive direction of the overlapped axes OX(O'X') (1) should read
>
>p'=m'u'. (2)

Okay. Still with ya.

>Less or more complicated derivations lead to the following transformation equations for momentum
>and mass (g(V)=1/sqrt(1-V^2/c^2)
>
>p=g(V)p'(1+V/u') (3)
>m=g(V)m'(1+Vu'/c^2) (4)

Okay.

>"Old fashioned" physicists say that (4) relates the "relativistic mass" m and the "relativistic mass m') of
>the same particle measured by observers from I and I' respectively.

I'm only 46. Can it really be said that I'm "Old fashioned"? :)

>If the particle is at rest in I' (u'=0)
>observers of that frame measure its "rest mass m(0) and (4) leads to
>
>m=g(V)m(0) (5)

Yup.

>Taking into account that c has the same magnitude in all inertial reference frames in relative motion, all
>transformation equations remain "relativistically correct" if we multiply both their sides by a power of c.
>Doing so with (4) we obtain nothing interesting because mc and m'c have no physical meaning (no
>tardyon can move with speed c).

The meaning of mc is that m is the relativistic mass times c. That is all.

>Multiplying both sides 0f (4) with c^2 leads to E=mc^2 and E'=m'c^2

So long as this is a isolated object then that's fine with me.

>respectively which has the physical dimensions of energy (4) becoming
>
>E=g(V)E'(1+Vu'/c^2)=g(V)(E'+Vp') (6)
>
>(3) becoming

p=g(V)(p'+VE'/c^2). (7)

E=g(V)E(0) (8) E(0)

>representing the rest energy.
>Presenting the transformation equations as (6), (7) and(8) the "new generation" of relativists have
>nothing to comment.

And this is what many physicists have been doing for years and years.

>Is the dispute between the generations solved? Are the frenzied debates motivated?

There will always be disputes Bernhard. Part of doing physics is using your imagination and people have different imaginations. It's as simple as that.

Pete

8. Aug 25, 2007

### daniel_i_l

In Spacetime Physics events are sperated by a timelike interval if it's possible for light to travel from one to the other. They're seperated by a spacelike interval if it's not possible.

9. Aug 25, 2007

### robphy

Of course, [assuming a nice spacetime] you mean that
"events are sperated by a lightlike (or null) interval if it's possible for light to travel from one to the other"
and
"events are sperated by a timelike interval if it's possible for some massive particle (travelling slower than light) to travel from one to the other"
and
"They're seperated by a spacelike interval if it's not possible for either light or some massive particle to travel from one to the other ."

10. Aug 25, 2007

### bernhard.rothenstein

Please tell me when do you consider that a space is nice (empty space?)
It is not suficient in the last sentence it is not sufficient to mention only light?
It is not advisable in all cases to invoque causality in the definition of timelike and space like?
Thanks

11. Aug 25, 2007

### robphy

Ok... here are some clarifications. (Initially, I was hoping to make the statements closer to the truth, rather than a thorough definition.

A nice spacetime is, loosely, one that has no causal anomalies [which would make the definition I am trying to use fail].

Yes, one should more correctly use a light-like particle [i.e., with zero rest-mass]... of which light is one example.

Actually, my preferred definition of Spacelike is "being orthogonal to timelike".
In Galilean spacetimes, with the lightcones opened up, being spacelike and being null coincide in the Galilean metric.
Null directions are probably better defined as eigenvectors of the boost transformations... this is true in both Galilean and Minkowskian spacetimes.
(These are described in my AAPT poster.)