today s paper modified gravity black holes and their
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Todays Paper: Modified Gravity Black Holes and Their Observable - PowerPoint PPT Presentation

Todays Paper: Modified Gravity Black Holes and Their Observable Shadows by John Mo ff at[arXiv:1502.01677] Elias Roland Most 6. June 2015 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 1 / 23 Galaxy Rotation Curves


  1. Today’s Paper: Modified Gravity Black Holes and Their Observable Shadows by John Mo ff at[arXiv:1502.01677] Elias Roland Most 6. June 2015 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 1 / 23

  2. Galaxy Rotation Curves Quick calculation µ v 2 r = G µ M r 2 r GM = ) v = r Most people assume this can be explained by a dark matter halo. Figure : M33 rotation curve [wikipedia:Galaxy Rotation Curve] Elias Roland Most Journal Club in High Energy Physics 6. June 2015 2 / 23

  3. But there are others.. Modified Newtonian Dynamics What if for very small accelerations Newtons second law was di ff erent, ✓ a ◆ F = m ⇠ a a 0 ✓ a ◆ 1 where ⇠ = 1 + a 0 a 0 a so that for small accelerations F = m a a 0 a ? Repeating the previous calculation 1 4 . = ) v = ( GMa 0 ) Elias Roland Most Journal Club in High Energy Physics 6. June 2015 3 / 23

  4. Modified Gravity An Overview Tries to explain galaxy rotation curves without dark matter. Modifies Newton’s constant G = G N (1 + ↵ ) Introduces scalar fields to mimic the running of couplings normally induced by an e ff ective theory of quantum gravity(e.g [arXiv:1202.2274]) Also introduces a vector field Elias Roland Most Journal Club in High Energy Physics 6. June 2015 4 / 23

  5. Modified Gravity S = S Grav + S � + S S + S M , where  1 1 d 4 x p� g � Z S Grav = G ( R + 2 Λ ) , 16 ⇡ Z d 4 x p� g  ✓ 1 4 B µ ⌫ B µ ⌫ + 1 ◆� 2 µ 2 � µ � µ + V ( � ) S � = � , ! and  1 Z d 4 x p� g ✓ 1 ◆ 2 g µ ⌫ r µ G r ⌫ G � V ( G ) S S = G 3 + 1 ✓ 1 ◆ 1 ✓ 1 ◆� 2 g µ ⌫ r µ ! r ⌫ ! � V ( ! ) 2 g µ ⌫ r µ µ r ⌫ µ � V ( µ ) + . µ 2 G G Elias Roland Most Journal Club in High Energy Physics 6. June 2015 5 / 23

  6. Two EoMs G µ ⌫ � g µ ⌫ Λ + Q µ ⌫ = 8 ⇡ GT µ ⌫ , where G µ ⌫ = R µ ⌫ � 1 2 g µ ⌫ R and Q µ ⌫ = G ( r ↵ r ↵ Θ g µ ⌫ � r µ r ⌫ Θ ) , where Θ ( x ) = 1 / G ( x ). r ⌫ B µ ⌫ + @ V ( � ) + 1 ! r ⌫ ! B µ ⌫ = � 1 ! J µ . @� µ Elias Roland Most Journal Club in High Energy Physics 6. June 2015 6 / 23

  7. Geodesic equations The test particle action is given by dx µ Z Z S TP = � m d ⌧ � � d ⌧!� µ d ⌧ . Then the geodesic equation takes the following form ✓ d 2 x µ dx ↵ dx � ◆ d ⌧ 2 + Γ µ = f µ , m ↵� d ⌧ d ⌧ where dx ⌫ dx ↵ � µ dx ↵ ✓ ◆ ✓ ◆ f µ = �! B µ ⌫ d ⌧ + � r µ ! � ↵ � � r ↵ ! . d ⌧ d ⌧ Obtaining equations of motions from this is tedious! Elias Roland Most Journal Club in High Energy Physics 6. June 2015 7 / 23

  8. Massive vector fields in QFT Detail may be found in [A.Zee: Quantum Field Theory in a Nutshell] Start from the action Z d 4 x { � µ D µ ⌫ � ⌫ + � µ J µ } , S = where D µ ⌫ = @ 2 + m 2 � g µ ⌫ � @ µ @ ⌫ . � D � µ e iS is just a Gaussian integral that can R The partition function Z = easily be solved to d 4 k (2 ⇡ ) 4 J µ ⇤ ( k ) � g µ ⌫ + k µ k ⌫ / m 2 W ( J ) = � i log Z = � 1 Z J ⌫ ( k ) . k 2 � m 2 + i " 2 Using that J is conserved, i.e. @ µ J µ = 0 d 4 k W ( J ) = 1 Z 1 (2 ⇡ ) 4 J µ ⇤ ( k ) k 2 � m 2 + i " J µ ( k ) 2 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 8 / 23

  9. Massive vector fields in QFT What is the force between two massive vector field point particles? Let J = J 1 + J 2 , where J i ( x ) = � (3) ( x � x i ). R d ke ikx , Neglecting self-interactions J i J i and using � ( x ) = e i ~ d 3 k k · ( ~ x 1 � ~ x 2 ) ✓Z ◆ Z d x 0 W ( J ) = k 2 + m 2 , (2 ⇡ ) 3 ~ = T e � mr 4 ⇡ r Elias Roland Most Journal Club in High Energy Physics 6. June 2015 9 / 23

  10. Massive vector fields in QFT ⌦ e � iHT ↵ Using that Z = we can identify E = i log Z / T = W / T = e � mr 4 ⇡ r . So the potential between two massive vector particles is of Yukawa type and leads to a repulsion. Note that this only applies in flat spacetime. Elias Roland Most Journal Club in High Energy Physics 6. June 2015 10 / 23

  11. Weak field limit To make predictions about galaxy rotation curves we need to consider the weak field limit g µ ⌫ = ⌘ µ ⌫ + h µ ⌫ . Then we would also have to perturb G , µ, � µ and things would get messy. Instead of doing this we can instead use our previous insights to guess x 0 � e � µ | ~ x � ~ x 0 | x 0 ) Z d 3 x 0 ⇢ ( ~ Z d 3 x 0 ⇢ � ~ Φ e ff ( ~ x ) = � G N (1 + ↵ ) x 0 | + G N ↵ x 0 | . | ~ x � ~ | ~ x � ~ Note that the prefactor of the second term has been chosen to cancel the the corrections at small distances, viz. e � mr ' 1, so that compatibility with Newtonian gravity is maintained. = ) This is indeed the result we would have obtained from the EOMs. Elias Roland Most Journal Club in High Energy Physics 6. June 2015 11 / 23

  12. Galaxy rotation curves revisited V(km/s) NGC3769 V(km/s) NGC3726 V(km/s) NGC3877 180 140 180 160 160 120 140 140 100 120 120 80 100 100 80 80 60 60 60 40 40 40 20 20 20 0 0 0 0 5 10 15 20 25 30 0 5 10 15 20 25 30 0 1 2 3 4 5 6 7 8 9 10 R(kpc) R(kpc) R(kpc) V(km/s) NGC3917 V(km/s) NGC3893 V(km/s) NGC3949 180 200 140 160 175 120 140 150 100 120 125 100 80 100 80 60 75 60 40 50 40 20 25 20 0 0 0 0 2.5 5 7.5 10 12.5 15 17.5 20 0 2 4 6 8 10 12 14 0 1 2 3 4 5 6 7 8 R(kpc) R(kpc) R(kpc) V(km/s) NGC3953 V(km/s) NGC3972 V(km/s) NGC4010 250 140 140 200 120 120 100 100 150 80 80 100 60 60 40 40 50 20 20 0 0 0 0 2 4 6 8 10 12 14 16 18 0 1 2 3 4 5 6 7 8 9 0 2 4 6 8 10 R(kpc) R(kpc) R(kpc) Figure 2. The best fit to the rotation velocity curves of the Ursa-Major sample. We fix α = 8 . 89 and µ = 0 . 042 kpc − 1 from the fits to the THINGS catalogue. We take the stellar mass-to-light ratio Υ � as the free degree of freedom. Figure : [arXiv:1306.6383] Elias Roland Most Journal Club in High Energy Physics 6. June 2015 12 / 23

  13. Galaxy rotation curves revisited These fits fix the values of ↵ and µ to ↵ = 8 . 89 µ = 0 . 042kpc � 1 As distances in the solar system are of the order of light minutes, these values are compatible with the Yukawa part of the gravitational potential being absent in solar system based experiments. Elias Roland Most Journal Club in High Energy Physics 6. June 2015 13 / 23

  14. Black hole solutions[arXiv:1412.5424] Assumptions: G = G N (1 + ↵ ) is constant ! = 1 µ is constant In this case the Einstein eq. reduces to R µ ⌫ = � 8 ⇡ GT � µ ⌫ , where T � µ ⌫ = � 1 µ B ⌫↵ � 1 � B ↵ 4 g µ ⌫ B ↵� B ↵� � 4 ⇡ Note that the Mass term / µ 2 has been neglected. Elias Roland Most Journal Club in High Energy Physics 6. June 2015 14 / 23

  15. Reissner-Nordstr¨ om solution This is nearly the same as for a charged black hole in standard GR, ◆ � 1 + GQ 2 + GQ 2 ✓ 1 � 2 GM ◆ ✓ 1 � 2 GM ds 2 = dt 2 � dr 2 � r 2 d Ω 2 . r 2 r 2 r r However, the charge is related to the mass by the following postulate p Q = ↵ G N M . Elias Roland Most Journal Club in High Energy Physics 6. June 2015 15 / 23

  16. Reissner-Nordstr¨ om The horizons are at p h i r ± = G N M 1 + ↵ ± 1 + ↵ r rS 7 6 5 r + 4 ������� r - 3 2 1 α 2 4 6 8 10 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 16 / 23

  17. Kerr solution As in GR we also have a Kerr-Newman type solution ⇢ 2 ( dt � a sin 2 ✓ d � ) 2 � sin 2 ✓ ⇢ 2 [( r 2 + a 2 ) d � � adt ] 2 � ⇢ 2 ds 2 = ∆ ∆ dr 2 � ⇢ 2 d ✓ 2 , where ∆ = r 2 � 2 GMr + a 2 + ↵ G N GM 2 , ⇢ 2 = r 2 + a 2 cos 2 ✓ . horizons s  a 2 � ↵ r ± = G N (1 + ↵ ) M 1 ± 1 � N (1 + ↵ ) 2 M 2 � G 2 1 + ↵ ergosphere s a 2 cos 2 ✓  � ↵ r E = G N (1 + ↵ ) M 1 + 1 � N (1 + ↵ ) 2 M 2 � . G 2 1 + ↵ Elias Roland Most Journal Club in High Energy Physics 6. June 2015 17 / 23

  18. Kerr solutions Horizons y rS 6 4 rE 2 r + �������� x r - rS - 6 - 4 - 2 2 4 6 rE ( GR ) - 2 - 4 - 6 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 18 / 23

  19. Photons in Reissner-Nordstr¨ om s r ph = 3 ✓ 8 ↵ ◆ 2 G N (1 + ↵ ) M 1 + 1 � . 9(1 + ↵ ) y rS 10 5 rPh �������� x rPh ( GR ) rS - 10 - 5 5 10 - 5 - 10 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 19 / 23

  20. Black hole shadow Figure : arXiv:1502.01677 Elias Roland Most Journal Club in High Energy Physics 6. June 2015 20 / 23

  21. Black hole shadow The black hole shadow is given by the set of all closed photon orbits. We have N (1 + ↵ ) M 2 � G N (1 + ↵ ) M ( r 2 � a 2 ) x = r ∆ + r ↵ G 2 , a [ r � G N (1 + ↵ ) M ] sin ✓ � 1 / 2 4 r 2 ∆ ⇢ [ r � G N (1 + ↵ ) M ] 2 � ( x + a sin ✓ ) 2 y = , where ∆ = r 2 � 2 G N (1 + ↵ ) Mr + a 2 + ↵ G 2 N (1 + ↵ ) M 2 . For a = 0 it is given as x 2 + y 2 p r shad ⌘ i 2 h p 3(1 + ↵ ) ± (9 + ↵ )(1 + ↵ ) = � 1 / 2 G N M . ⇢ i 2 h p 4 (1 + ↵ ) ± (9 + ↵ )(1 + ↵ ) � 16(1 + ↵ ) Elias Roland Most Journal Club in High Energy Physics 6. June 2015 21 / 23

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