Relative to what do you measure c?

[Nacelunk]

Suppose you travel at 0.9999999c in a spaceship and you turn on a flashlight forwards. As far as I know speeda light is always constant 3*10e8. So it will fly out at that speed. But relative to what - the spaceship or our inertial frame of reference (outside the ship)? And i have another, a little off topic question. When you fly at speed close to speeda light, you notice that outside observer's (at rest) clock is running slow. The outside observer also notices your clock running slow, right? I don't really get that.

Sorry, for offtopic again but don't wanna create three different threads:

concerning equation E=mc*2. I don't get why speed of light affects amount of energy matter has. I don't see the relation at all. If c was 2 times greater, would the amount energy, therefore, increase by 4 times? If so, why?

[swansont]

Suppose you travel at 0.9999999c in a spaceship and you turn on a flashlight forwards. As far as I know speeda light is always constant 3*10e8. So it will fly out at that speed. But relative to what - the spaceship or our inertial frame of reference (outside the ship)? And i have another, a little off topic question. When you fly at speed close to speeda light, you notice that outside observer's (at rest) clock is running slow. The outside observer also notices your clock running slow, right? I don't really get that.

Sorry, for offtopic again but don't wanna create three different threads:

concerning equation E=mc*2. I don't get why speed of light affects amount of energy matter has. I don't see the relation at all. If c was 2 times greater, would the amount energy, therefore, increase by 4 times? If so, why?

 

Both frames will measure c to be the same. They will disagree on length and time measurements. The disagreements are symmetric to the two reference frames, because both can claim to be at rest — neither one has any measurement that would prove the other one to be incorrect. It's all a consequence of c being constant, and takes a little getting used to.

[rajama]

Suppose you travel at 0.9999999c in a spaceship and you turn on a flashlight forwards. As far as I know speeda light is always constant 3*10e8. So it will fly out at that speed. But relative to what - the spaceship or our inertial frame of reference (outside the ship)? And i have another, a little off topic question. When you fly at speed close to speeda light, you notice that outside observer's (at rest) clock is running slow. The outside observer also notices your clock running slow, right? I don't really get that.

Sorry, for offtopic again but don't wanna create three different threads:

concerning equation E=mc*2. I don't get why speed of light affects amount of energy matter has. I don't see the relation at all. If c was 2 times greater, would the amount energy, therefore, increase by 4 times? If so, why?

 

On your third topic: if time rate slows isn’t it reasonable for mass to also increase – to balance things out?

 

And of course this would be a reciprocal arrangement: if your spaceship slams into Earth, it doesn’t matter whether you’re on the ship or standing on the ground, the bang should be equally energetic…

[TriggerGrinn]

About velocity and squaring. Also Energy and Velocity:

 

When you refer to energy of an object you are talking about how much work it can do before it stops relative to the reciever.

 

Somewhere between zero energy(rest) and some energy (motion) there is two absolutely different states occuring, you either got motion or you don't.

 

Now, when we consider that, any moving object already contains energy. (x) kinetic energy. So therefore if you double its speed, you have added energy to it that it already had. So at double the speed it has double the energy previous + the previous energy + the slower previous energy of that + plust the slower energy of that + so on and so forth untill its at 'rest'.

 

Because it deals with the 'stopping' of things... we end up required to square velocities, and this is worked out through as I understand

 

[math] work = fd = mad = m a v' t [/math] = [math]m a (\frac {V_0 + V_f}{2}) t = m \frac {V_f}{t} \frac {V_f}{2}t = \frac {1}{2} mv^2 = K_e = Kinetic Energy[/math]

 

which is also, and may help for you understanding:

 

[math]K_e = \frac {p^2}{2m} = \frac{(m*v)(m*v)}{m+m}[/math]

 

 

However, for mass, a massive object can only deliver a limited value of energy to another massive object in space, and that limited value is entirely dependant on the particular velocity.

That particular 'value' is figured out as:

[math]\gamma = \frac {1}{\sqrt {1- \frac{v^2}{c^2}}}[/math]

or

 

but we don't need to get into that as it gets too complex.

 

 

The object that gets hit must be equal to or greater than the moving object in order for full momentum to be exchanged, in variable free conditions. (variable free, meaning IE: perfect billiard balls)

 

Which directly goes on to mean that full energy can not be extracted from objects, for a relative observer, if the observers frames mass is considerably less than the moving object with low velocity.

 

That may have sounded confusing, but think about this.

 

On earth we use friction attatched to the earth and when we attatch ourself to the earth we make earth and ourselves OUR frame of rest. The mass of the earth is so great, it does not move noticelably from a small moving object, so a small moving object can attain a high amount of energy, at relatively low velocity.

 

Now lets see what velocity has to do with this.

 

However if you consider yourself in space and an object comes at you near the speed of light, and when it whizzes right beside you, you hook up a perfect elastic band that is attatched to your frame. Lets consider both your masses are equal. You will be able to get a certain amount of energy out of it, but as you use it, you are also getting pulled along with that object and in result, stripping velocity away from the object as you both begin to chug along, one losing velocity while the other gains velocity, and both effecting the capable energy to be extracted. As you see the faster you slow down the object, the greater the energy its going to deliver to your frame, before you being moving. And of course, the slower it slows down, the less energy you are going to get out of it, because of the more time involved in the energy exchange. Your inertia is tied in with the rate of acceleration. The 'time' it takes to get to [math]V_f[/math] velocity final.

 

 

Now if we exclude the elastic band and have two objects collide, we see that when things collide the time of momentum exchange is nearly instantanious. So the less time invovled the greater the energy that can be put to use.

 

Now finally, we move onto light. It has no mass. So for light, when it exchanges energy, it does it so quickly, that is, a high velocity, the fastest velocity [math]c[/math] that all energy can be put into use, as described before. So We end up with

[math]K_e=\frac {1}{2} mv^2[/math] turning into [math]E=mv^2[/math] where V happens to be the speed of light, and this is symbolised as C. ANd finally we look at [math]E = MC^2[/math]

[TriggerGrinn]

But, don't trust me on the mathmatics just yet, someone with better experience, will have to double check my work.

[swansont]

Muddled and so difficult to always say exactly where it strayed. However

 

The object that gets hit must be equal to or greater than the moving object in order for full momentum to be exchanged, in variable free conditions. (variable free, meaning IE: perfect billiard balls)

 

Momentum transfer will be maxmized when the masses are equal

 

Now finally, we move onto light. It has no mass. So for light, when it exchanges energy, it does it so quickly, that is, a high velocity, the fastest velocity [math]c[/math] that all energy can be put into use, as described before. So We end up with

[math]K_e=\frac {1}{2} mv^2[/math] turning into [math]E=mv^2[/math] where V happens to be the speed of light, and this is symbolised as C. ANd finally we look at [math]E = MC^2[/math]

 

[math]K_e=\frac {1}{2} mv^2[/math] doesn't "turn into" [math]E=mv^2[/math]

 

[math]\frac {1}{2} mv^2[/math] is a nonrelativistic approximation, valid for v<<c

[spikerz66]

on your third point: i fould a great explanation on a website on why the speed of light is used in the equation

 

http://www.osti.gov/accomplishments/nuggets/einstein/speedoflight.html

use this link

[TriggerGrinn]

Yah thanks I am no pro at organizing the mathmatics..

 

but I will work to back up the verbal statements