# Speed vector

1. Sep 10, 2006

### alpaolo

is a delirium?:yuck:

1) The photons was influenced by gravity?
2) I think yes, but the light don't lies on earth.
3)The light haves a energy wich opposed to gravity.
4)A body with energy in opposite sense of gravity don't feel a speed of a container ( a jump in elevator or other)
5)The newton's speed vector formula is applied when a mass lies in a body with a specified speed.
6)The light is'nt dragged by a earth
7)Is'nt possible to apply the speed vector formula to the light, but the formula is correct only if applied to body wich don't opposed to the gravity.
8) The finish speed of the light is not a paradox and is not in opposite to newton's formula.

Sorry for the english!

2. Sep 10, 2006

### HallsofIvy

Staff Emeritus
Yes, general relativity predicts that and it was proved by the famous "eclipse" experiment (of 1928?).

What does "lying on earth" have to do with the question?

Gravity is a force. There is no such thing as an "energy" opposing a "force".

Again, you are confusing "energy" with "force".

Mass lying in a body? Are you talking about a mass with a given constant speed? No, Newton's laws do not apply only to constant speed.

Not sure what you mean by "dragged" but the path of light surely is affected by a massive body such as the earth. (Although the earth isn't all that massive and the effect is very small.)

Which "speed vector formula" are you talking about? We use Newton's formulas in gravitational situations all the time.

I don't know that anyone ever said it was!

Very good English! Better than my (put whatever language you like here).

3. Sep 10, 2006

### alpaolo

speed of light

I try to be more precise.

I wrote this because the Scientist says "Is impossible to sum the speed of a train in wich there is a light generator ( like any other body) and a light speed because the speed of light is costant".
I think, the speed light is costant because there are'nt any possibility to drag the light and also verify it.
Infact only a moving body (es A) that is lean in an another moving body like a train(B) can be view by a observer with a speed that is a sum of 2 speed(Va+vb).
If a bird ( like a colibri) uses its energy to lift in a middle height (contrast the gravity)but don't move in any direction respect the observer, the train moves under the bird and the speed is not summed; if the bird move in some direction with a certain speed only this speed is perceived by an observer not the sum.

The light, ( excuse for the relation) light a bird stay in suspension (referred to the observer), in this case the only speed perceiced is a light speed and no more.

4. Sep 10, 2006

### HallsofIvy

Staff Emeritus
What scientist is this? Are you referring to someone on this forum using the name "Scientist" or scientists in general? If the former, I can't speak to that. If the latter, it's simply not true. You definitely can "sum" the speed of light and another object (such as a train)- you just have to use the correct formula. If a train is moving at speed v (relative to the ground) and I walk at speed u (relative to the train) toward the front of the train, then my speed (relative to the ground) is
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
If both my and the train's speeds are small compared to c, then that will be indistinguishable from u+ v so we can use u+ v as a simpler, approximate formula. If either or both are a large fraction of c, then that is very different from u+ v. In particular if u= c, for example if we shine a light forward from at train going at speed v, then we have
$$\frac{v+ c}{1+ \frac{vc}{c^2}}= \frac{v+c}{1+ \frac{v}{c}}$$
Multiplying both numerator and denominator by c:
$$\frac{(v+c)c}{c+v}= c$$.
We have "summed" the speeds but using the correct formula just gives c.

I still don't know what you mean by "drag the light". Certainly, light has been observed having its velocity altered by gravity although its speed is not.

All you are saying is that if the bird's speed, relative to the train, is 0, there is no sum: that's true, or rather the sum is v, whether you use v+ 0= v or $\frac{v+0}{1+ 0v/c^2}= v$.
If your observer is standing on the ground watching the train, and bird, go by then he is observing the "sum" of their speeds.

??Stays in suspension?? What do you mean by that? I assumed before that you were saying the bird was stationary in the air relative to the train. Certainly no observer would see light acting like that! And is your observer on the train or on the ground?

5. Sep 10, 2006

### alpaolo

Thank you HallsOfIvy, I use a strange word because I don't know the exact word for illustrated my thinking.
If the bird speed is 0 relative to train the bird speed relative to the ground is a train speed. Yes.
I say other: The top of the train is open, the train is very long the train speed is v1 relative to the ground.
The bird spedd is v2 relative to the ground, the bird like a elevator go down inside the train( no air effects was supposed ). The observer view the bird at speed v2.
The bird go down again the lateral speed is always v2, but the bird touch the floor immediately the bird speed viewing by observer is V1+V2.
When the bird stopping to contrast the gravity the speeds summing.
Yes is natural but this idea is very strange....
There's a formula that consider this? i.e. adding something to the Newton's formul that consider this?

I know this idea can appear crazy....

Is corret thinking the light always stationary respect a train or earth or anithing in for example?