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Light and Time... doesn't match

Light is emitted by matter which is at a speed lower than \(c\). Also, Light travels only at a speed \(c=299792458\) m/s. Also, Acceleration \(a=\frac{\Delta v}{\Delta t}\). as Light can only travel at \(c\), light must achieve \(c\) in \(\Delta t=0\). if \(a=\frac{\Delta v}{0}\) then \(a=\infty\).Will it happen in this universe? or is there any medium in this universe? I request everyone to comment on this... Please Don't Ignore... Even if yours or my answer is wrong,comment on this...

Note by Fahad Shihab
3 years, 6 months ago

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Your assumption that \(\Delta t=0\) is right but you also assume that \(\Delta v \neq 0\) which is not right. Actually \(\Delta v\) is also equal to \(0\). This is because the initial and final speed are same and equal to \(c\). Shubham Srivastava · 3 years, 5 months ago

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@Shubham Srivastava how? Light is emitted by matter which is travelling at speeds below \(c\). So, if it wants to go at \(c\), a change of velocity must occur. so,\(\Delta v\neq0\). Fahad Shihab · 3 years, 5 months ago

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\(c\), is the speed of light in a vaccum, light doesn't slow down (de-accelerate), it just travels slower in a different medium. End of story. Yong See Foo · 3 years, 5 months ago

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@Yong See Foo i am talking about light in a vacuum. Not mediums. Fahad Shihab · 3 years, 5 months ago

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Now another news... check this out! Fahad Shihab · 3 years, 5 months ago

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Photons can only travel at a speed of \(c\) in which beforehand, for instance when emitted from a atom; they simply didn't exist in the form of a photon. Photons theoretically have no mass as they do not interact with the Higgs field; hence cannot be accelerated. Joel Jablonski · 3 years, 5 months ago

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@Joel Jablonski OK. so, if it didn't exist in the form of a photon, it must have a travelling speed, that is \(c=299792458ms^{-1}\) and according to my knowledge, \(a=\frac{\Delta v}{\Delta t}\) as acceleration is a funcion of time, not mass. So, even if its mass is zero, it should have an acceleration. Can't it be? Fahad Shihab · 3 years, 5 months ago

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@Fahad Shihab I understand where you are coming from and I agree that acceleration is a function of the time, but it is only a representation. When the time 0 the velocity is already c and it did not start from a velocity of 0, it simply didn't exist in the form of a photon as I previously said. The travelling speed that you are talking about was just a form of energy that was suddenly converted into a photon, it didn't exactly have a 'speed' more like a potential.

The acceleration of a particle is not determined by the time, it is determined by its mass or the Higgs field. I haven't studied general relativity in depth though seeming as the photon does not interact with the Higgs field, it can not change its speed. Hence, the acceleration of the photon is \(a=\frac{Δv}{Δt}\) where change in velocity is 0, hence acceleration must always be 0 no matter what time, even at t=0. Joel Jablonski · 3 years, 5 months ago

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@Joel Jablonski So, physically, light isn't visible to us. That means it is a wave and photons are imaginary. Then what is the reason for self propagation of light? Fahad Shihab · 3 years, 5 months ago

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Rule of Mathematics, In Shape \(\frac{a}{b}\) , with b \(\neq\) 0, so there exists this possibility ! \(\Delta\)\(t\) \(\neq\) 0 Gabriel Merces · 3 years, 5 months ago

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@Gabriel Merces in relativity, it is said that light can't travel at a speed lower than \(c\) Fahad Shihab · 3 years, 5 months ago

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@Fahad Shihab It has to be a time for less than Be, After All Denominator must be \(\neq\) of 0 Alex Dias · 3 years, 5 months ago

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@Alex Dias that's why i am saying that even light can travel at speeds lower than \(c\) Fahad Shihab · 3 years, 5 months ago

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